Radius of Curvature when dy/dx is undefined

This has nothing to do with the particulars of the function that you applied it to, this is solely a feature of the radius of curvature formula itself.

To say this with extreme brevity: the formula for $\rho$ is invariant when you reflect across the line $y=x$.

To explain this in greater length, starting from the radius of curvature formula for a function $y=f(x)$ $$\rho = \left( 1 + \left(\frac{dy}{dx}\right)^2 \right)^{3/2} \biggm/ \frac{d^2y}{dx^2} $$ let's imagine for the moment that we are at a point $(x,y)$ where $\frac{dy}{dx} \ne 0$. The inverse function theorem therefore applies at this point, and so we have an inverse function $f^{-1}(y) = x$, and we know its derivative $(f^{-1})'(y) = \frac{1}{f'(x)}$.

This allows us to make some substitutions.

First, in place of $\frac{dy}{dx}$ one can substitute $$\frac{1}{(f^{-1})'(y)} = \frac{1}{dx/dy} $$

Second, one can also work out a formula for the second derivative of the inverse function: $$\frac{d^2 x}{dy^2} = - \frac{d^2 y/ dx^2}{(dy/dx)^3} $$ and so in place of $\frac{d^2y}{dx^2}$ one can substitute $$-\frac{d^2x}{dy^2} (dy/dx)^3 = - \frac{d^2x}{dy^2}\frac{1}{(dx/dy)^3} $$ When you make these substitutions you get $$\rho = \left( 1 + \frac{1}{(dx/dy)^2} \right)^{3/2} \biggm/ \left(- \frac{d^2x}{dy^2}\frac{1}{(dx/dy)^3}\right) = - \left( 1 + \left(\frac{dx}{dy}\right)^2 \right)^{3/2} \biggm/ \frac{d^2x}{dy^2} $$ Now there's one piece of funny business going on here, which is that pesky minus sign. Well, all that's happening here is that $\rho$ is really a non-negative quantity, and the original formula was missing absolute value symbols around the whole thing.

Okay, that's all fine at a point $(x,y)$ where $\frac{dy}{dx} \ne 0$; and at such a point we should notice that $\frac{dx}{dy} \ne 0$.

What about a point $(x,y)$ where $\frac{dy}{dx} = 0$ and $\frac{dx}{dy}$ is undefined? Well, we just use the $\frac{dy}{dx}$ formula anyway.

And what about a point $(x,y)$ where $\frac{dy}{dx}$ is undefined and $\frac{dx}{dy}=0$? In that case, we use the $\frac{dx}{dy}$ formula. If that still makes you unhappy, then you can make two variable substitutions. First, substitute $s=x$, $t=y$ and rewrite the formulas. And when you've done that, then substitute $y=s$ and $x=t$ and rewrite the formulas again, and out will pop the same old formula for $\rho$.

(But but but... what about a point where neither $\frac{dy}{dx}$ and $\frac{dx}{dy}$ are undefined? Well, now you are out of luck. The radius of curvature is not defined at such a point.)

There is one other thing to say here.

What is going on under the hood is that the substitution $y'=x$, $x'=y$, which leaves the formula for $\rho$ unchanged (if you are careful with absolute values), represents a rigid motion of the plane namely a reflection across the line $y=x$.

But there are other interesting rigid motions of the plane. For instance, translations to left or right given by $x'=x+a$, or translations up or down given by $y'=y+b$; each of those substitutions leaves the formula for $\rho$ unchanged.

And there's one more REALLY interesting rigid motion to ponder, namely a rotation of the plane through an angle of $\theta$, given by the formulas $$x' = \cos\theta \, x - \sin\theta \, y $$ $$y' = \sin\theta \, x + \cos\theta \, y $$ If you go through the whole painful procedure of working out how this coordinate change affects the formula for curvature $\rho$, PRESTO! It is also unchanged.

The point here is that the radius of curvature $\rho$ is really a geometric measurement of a graph, independent of the coordinate system as long as you change coordinates by a rigid motion.

The radius of curvature $\rho=\frac1{\kappa}$ is a geometric invariant of the curve and is independent of which (regular) parametrisation one uses. At the particular point you mentioned, while the curve itself is smooth, it is not represented as the graph of a smooth function over the $x$-axis. Therefore one needs to use another parametrisation, for example as the graph over the $y$-axis.

Recall that in parametric form, where $x$ and $y$ are functions of $t$, the radius of the tangent circle is:

$$\rho = \left|\frac{((x')^2 + (y')^2)^{3/2}}{x'y'' - x''y'}\right|$$

Note that this formula is symmetric in $x$ and $y$. Which makes logical sense geometrically, because a circle doesn't change its radius if reflected across the line $y = x$. Nor does a curve change its curvature. It's the same thing, just viewed from a different perspective.

If $x = t$, then $x' = 1$ and $x'' = 0$, so the formula simplifies to:

$$\rho = \left|\frac{(1 + (y')^2)^{3/2}}{y''}\right|$$

Or in Leibniz notation, noting that $dt = dx$,

$$\rho = \left|\frac{(1 + (\frac{dy}{dx})^2)^{3/2}}{\frac{d^2y}{dx^2}}\right|$$

Alternatively, if $y = t$, then $y' = 1$ and $y'' = 0$, so:

$$\rho = \left|\frac{((x')^2 + 1)^{3/2}}{-x''}\right| = \left|\frac{(1 + (\frac{dx}{dy})^2)^{3/2}}{\frac{d^2x}{dy^2}}\right|$$

We can show that $dy/dx$ is undefined for $\rho$ while $dx/dy$ will give the appropriate $\rho$ using implicit differentiation. In each case, we'll let $y'$ to stand in for $dy/dx$ and $x'$ to stand in for $dx/dy$.

Implicit differentiation in terms of $x$...

$$2y y' = 3x^2 \\ y' = \frac {3x^2} {2y} \\ (y')^2 = \frac {9x^4}{4y^2}$$

To find $y''$, we have (after using the quotient rule and simplifying) we get $$\frac {2y (6x) - 3x^2 (2 y')}{4y^2} \\ \\ \frac {3x (y-xy')}{2y^2} \\$$ Subsituting $y'$ and simplifying we get for $y''$ $$\frac{6xy^2-9x^4}{4y^3}$$ Now we go to the original formula and fill things in...

$$\rho = \frac {(1+(y')^2)^{3/2}}{y''} \rightarrow \frac {(1+(\frac {9x^4}{4y^2}))^{3/2}}{\frac{6xy^2-9x^4}{4y^3}}$$

However, there's one problem. Since $y=0$, the denominators would be $0$; thus $\rho$ is not defined.

Implicit differentiation in terms of $y$...

$$ 2y = 3x^2 x' \\ x' = \frac {2y}{3x^2} \\ (x')^2 = \frac {4y^2}{9x^4}$$

Again, to find $x''$ (using the quotient rule and simplifying) we have $$\frac {3x^2 (2) - 2y (6x x')}{9x^4} \\ \frac {2x (x - 2y x')}{3x^4} \\$$ Subsituting $x'$ and simplifying we get for $x''$ $$\frac{6x^3-8y^2}{9x^5}$$

Now we go to the original formula and fill things in again...

$$\rho = \frac {(1+(x')^2)^{3/2}}{x''} \rightarrow \frac {(1+(\frac {4y^2}{9x^4}))^{3/2}}{\frac{6x^3-8y^2}{9x^5}}$$

Now we have something much nicer...since $y=0$, the $y$ terms vanish, leaving us with the $x$ terms in which we can substitute $x=-2$. Then we get $$\rho= \frac {1} {\frac {6(-2^3)}{9(-2^5)}} = \frac {9}{6} \cdot 4 = 6$$.

Note: We can also do each explicitly, first with $y = \pm \sqrt {(x^3 +8)}$ and then with $x=\sqrt[3]{(y^2-8)}$, but the work is much longer. Also, we run into problems with $y = \pm \sqrt {(x^3 +8)}$ because $x$ is not defined for $x < -2$. However, when we define in terms of $x$, $x=\sqrt[3]{(y^2-8)}$ is valid for all real numbers.