If a x^2+2 h x y+b y^2=3 , then (d^2 y)/(d x^2)=

We have, a x²+2 h x y+b y²=3 On differentiating with respect to $x$, we get $$\begin{aligned}& 2 a x+2 h x \frac{d y}{d x}+2 h y+2 b y \frac{d y}{d x}=0 \\ & \frac{d y}{d x}=-\frac{(a x+h y)}{(b y+h x)} \end{aligned}$$Again differentiating with respect to $x$, we get $$\begin{aligned}& 2 a+2 h x \frac{d^2 y}{d x^2}+2 h \frac{d y}{d x}+2 h \frac{d y}{d x}+2 b\left(\frac{d y}{d x}\right)^2 \\ \text{ and } +2 b y \frac{d^2 y}{d x^2}=0 \end{aligned}$$$$\begin{aligned} \text{ and } \frac{d^2 y}{d x^2}(h x+b y)=-\left(a+2 h \frac{d y}{d x}+b\left(\frac{d y}{d x}\right)^2\right) \\ \text{ and } \frac{d^2 y}{d x^2}(h x+b y)= \\ \text{ and } -\left(a+2 h\left(\frac{-a x-h y}{h x+b y}\right)+b\left(\frac{-a x-h y}{h x+b y}\right)^2\right) \end{aligned}$$\frac{d^2 y}{d x^2}(h x+b y)=-\left[\frac{a(h x+b y)^2-2 h(a x+h y)(h x+b y)+b(a x+h y)^2}{(h x+b y)^2}\right] \frac{d^2 y}{d x^2}=\frac{3\left(h^2-a b\right)}{(h x+b y)^3}