Given, $a{x}^2+2hxy+b{y}^2=0$

Now, differentiating both the sides with respect to $x$

$\Rightarrow 2ax+2h(y+x\frac{dy}{dx})+2by\frac{dy}{dx}=0$

$\Rightarrow 2hx\frac{dy}{dx}+2by\frac{dy}{dx}+2ax+2hy=0$

$\Rightarrow 2\frac{dy}{dx}(hx+by)=-2(ax+hy)$

$\Rightarrow \frac{dy}{dx}=-\frac{\left(ax+hy\right)}{hx+by} ...\left(1\right)$

Now, again differentiating with respect to $x$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\left[\frac{\left(a+h{\displaystyle \frac{dy}{dx}}\right)(hx+by)-\left(h+b{\displaystyle \frac{dy}{dx}}\right)(ax+hy)}{(hx+by{)}^2}\right]$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{\left[\left(ahx+h^2{\displaystyle \frac{dy}{dx}}x+aby+bh{\displaystyle \frac{dy}{dx}}y\right)-\left(ahx+ab{\displaystyle \frac{dy}{dx}}x+h^{2}y+hby{\displaystyle \frac{dy}{dx}}\right)\right]}{(hx+by{)}^2}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\left[\frac{(ab-h^2)y+(h^2-ab)x{\displaystyle \frac{dy}{dx}}}{(hx+by{)}^2}\right]$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-(ab-h^2)\left[\frac{y-x{\displaystyle \frac{dy}{dx}}}{(hx+by{)}^2}\right]$

Now, put $\frac{dy}{dx}=-\frac{\left(ax+hy\right)}{(hx+by)}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\left(ab-h^2\right)\left[\frac{y-x\left(-{\displaystyle \frac{\left(ax+hy\right)}{(hx+by)}}\right)}{(hx+by{)}^2}\right]$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\left(ab-h^2\right)\left[\frac{a{x}^2+2hxy+b{y}^2}{(hx+by{)}^2}\right]$

As given $a{x}^2+2hxy+b{y}^2=0$

So, $\frac{d^{2}y}{d{x}^2}=0$