Problem 1, Ch. 6 in Piskunov's, Differential and Integral calculus 
Find the curvature of the curve at indicated points
$b^2x^2+a^2y^2=a^2b^2$ at $(0,b)$ and $(a,0)$

My attempt
$\displaystyle{\kappa=\frac{|\frac{d^2{y}}{dx^2}|}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}}$
Differentiating the implicit equation with respect to $x$,
$2b^2x+2a^2yy'=0\\
y'=-\frac{b^2}{a^2}{\cdot}\frac{x}{y}$
Differentiating again with respect to x,
$y''=-\frac{b^2}{a^2}{\cdot}\frac{y-xy'}{y^2}$
a) At $(0,b)$
$y'=0\\
y''=-\frac{b^2}{a^2}{\cdot}\frac{b-(0)(0)}{b^2}=-\frac{b}{a^2}$
$\displaystyle{\kappa=\frac{|-\frac{b}{a^2}|}{\left[1+\left(0\right)^2\right]^\frac{3}{2}}=\frac{b}{a^2}}$
b) At $(a,0)$
$y'=\infty\\
y''=-\frac{b^2}{a^2}{\cdot}\frac{\frac{y}{y'}-x}{\frac{y^2}{y'}}=\infty$
I am not sure, if my solution to part (b) is correct. The book says, the answer must be $\frac{a}{b^2}$.
 A: Alternatively, observe that the given equation can be parametrized as follows:  $$(x(t), y(t)) = (a \cos t, b \sin t).$$  Then the curvature is given by $$\kappa(t) = \frac{|x'(t) y''(t) - x''(t) y'(t)|}{((x'(t))^2 + (y'(t))^2)^{3/2}}.$$  We compute $$(x'(t), y'(t)) = (-a \sin t, b \cos t), \\ (x''(t), y''(t)) = (-a \cos t, -b \sin t);$$ hence $$\kappa(t) = \frac{|ab|}{(a^2 \sin^2 t + b^2 \cos^2 t)^{3/2}}.$$ Since the choice $t = \pi/2$ corresponds to $(0,b)$, and the choice $t = 0$ corresponds to $(a,0)$, it is now trivial to obtain $\kappa(0) = |ab|/b^3 = a/b^2$, and $\kappa(\pi/2) = |ab|/a^3 = b/a^2$, for $a, b > 0$.
A: we get the curvature $$\displaystyle{\kappa=\frac{|\frac{d^2{y}}{dx^2}|}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}}$$
now $y'=-\frac{b^2}{a^2}{\cdot}\frac{x}{y}$
$y''=\frac{-a^2y'^2-b^2}{a^2y}$
putting up the values in formula 
we get 
$$\kappa =\frac{\frac{a^2y'^2+b^2}{a^2y}}{\bigg[1+\frac{b^4}{a^4}{\cdot}\frac{x^2}{y^2}\bigg]^\frac{3}{2}} $$
$$\kappa =\frac{\frac{b^4x^2+a^2y^2b^2}{a^2y^3}}{\frac{\bigg[a^4y^2+b^4{\cdot}x^2\bigg]^\frac{3}{2}}{a^2y^3}} $$
now it can be solved
A: Use chain rule for double differentiation. 
I think you should multiply by dy/dx in the expression for y''. 
For example,let y=1/x 
So y"=2/x^3
But the way you have done would give y"=-2y=-2/x
A: I solved it as follows:
$\begin{align}
y'&=-\frac{b^2}{a^2}\cdot\frac{x}{y}\\
y''&=-\frac{b^2}{a^2}\cdot{\frac{y-xy'}{y^2}}\\
&= -\frac{b^2}{a^2}\cdot{\frac{y-x\cdot{\left(-\frac{b^2}{a^2}\cdot\frac{x}{y}\right)}}{y^2}}\\
&=-\frac{b^2}{a^2}\cdot{\frac{a^{2}y^{2}+b^{2}x^{2}}{a^{2}y^{3}}}\\
&=-\frac{b^2}{a^2}\cdot{\frac{a^{2}b^{2}}{a^{2}y^{3}}}\\
&=-\frac{b^{4}}{a^{2}}\cdot{\frac{1}{y^{3}}}
\end{align}$
The radius of curvature $\kappa$ is given by,
$\begin{align}
\kappa&=\frac{|y''|}{[1+y'^{2}]^\frac{3}{2}}\\
&=\frac{|-\frac{b^{4}}{a^{2}}\cdot{\frac{1}{y^{3}}}|}{\left[1+\left(-\frac{b^2}{a^2}\cdot\frac{x}{y}\right)^{2}\right]^\frac{3}{2}}\\
&=\frac{a^{4}b^{4}}{\left[a^{4}y^{2}+b^{4}x^{2}\right]^\frac{3}{2}}
\end{align}$
