We have,
$$
a x^2+2 h x y+b y^2=3
$$

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
$$
\begin{aligned}
& 2 a+2 h x \frac{d^2 y}{d x^2}+2 h \frac{d y}{d x}+2 h \frac{d y}{d x}+2 b\left(\frac{d y}{d x}\right)^2 \\
&+2 b y \frac{d^2 y}{d x^2}=0
\end{aligned}
$$
$$
\begin{aligned}
& \frac{d^2 y}{d x^2}(h x+b y)=-\left(a+2 h \frac{d y}{d x}+b\left(\frac{d y}{d x}\right)^2\right) \\
& \frac{d^2 y}{d x^2}(h x+b y)= \\
&-\left(a+2 h\left(\frac{-a x-h y}{h x+b y}\right)+b\left(\frac{-a x-h y}{h x+b y}\right)^2\right)
\end{aligned}
$$
$\frac{d^2 y}{d x^2}(h x+b y)=-$
$$
\left[\frac{a(h x+b y)^2-2 h(a x+h y)(h x+b y)+b(a x+h y)^2}{(h x+b y)^2}\right]
$$
$$
\frac{d^2 y}{d x^2}=\frac{3\left(h^2-a b\right)}{(h x+b y)^3}
$$