If $ax^2 + 2hxy + by^2 = 1$, prove that $\frac{d^2y}{dx^2} = \frac{h^2-ab}{(hx + by)^3}$.
I did try to simplify it, by taking derivatives on both the sides, and taking the $\frac{dy}{dx}$ common, and applying the quotient rule.
The answer I get is $$\frac{d^2y}{dx^2} = 4ha + \frac{2ab}{(2h+b)^2}.$$
Edit 1: Sorry, it's now $4ha/(2h+b)^3$. I forgot to take the 2nd derivative the first time. But it's still not the answer.
\begin{align*} \frac{d}{dx}\left(ax^2+2hxy+by^2\right) & =0\\ 2ax+ 2h\underbrace{\left(x \, \frac{dy}{dx}+y\right)}_{\text{product rule}} + 2by\frac{dy}{dx}&=0\\ \frac{dy}{dx}&=-\frac{ax+hy}{hx+by}\\ \frac{d^2y}{dx^2}&=-\left[\frac{(hx+by)\frac{d}{dx}(ax+hy)-(ax+hy)\frac{d}{dx}(hx+by)}{(hx+by)^2}\right]\\ \frac{d^2y}{dx^2}&=-\left[\frac{(hx+by)\left(a+h\frac{dy}{dx}\right)-(ax+hy)\left(h+b\frac{dy}{dx}\right)}{(hx+by)^2}\right]\\ \frac{d^2y}{dx^2}&=-\left[\frac{\left(h^2x-abx\right)\frac{dy}{dx}+aby-h^2y}{(hx+by)^2}\right]\\ \frac{d^2y}{dx^2}&=-(h^2-ab)\left[\frac{x\frac{dy}{dx}-y}{(hx+by)^2}\right]\\ \frac{d^2y}{dx^2}&=-(h^2-ab)\left[\frac{-x\left(\frac{ax+hy}{hx+by}\right)-y}{(hx+by)^2}\right]\\ \frac{d^2y}{dx^2}&=(h^2-ab)\left[\frac{\color{red}{ax^2+2hxy+by^2}}{(hx+by)^3}\right]\\ \frac{d^2y}{dx^2}&=(h^2-ab)\left[\frac{\color{red}{1}}{(hx+by)^3}\right] \end{align*}