It is given that,

$a\ne b,x\ne n\pi ,n\in z$ and $y^2=a^2{\cos }^{2}x+b^2{\sin }^{2}x$

Differentiate the above equation w.r.t. $x$,

$2y\frac{dy}{dx}=-a^2\sin 2x+b^2\sin 2x$

$2y\frac{dy}{dx}=\left(b^2-a^2\right)\sin 2x$

Again differentiate the above equation w.r.t. $x$,

$2\left(y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2\right)=2\left(b^2-a^2\right)\cos 2x$

$y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=\left(b^2-a^2\right)\cos 2x$

Multiply $y^2$ on both sides,

$y^3\frac{d^{2}y}{d{x}^2}+y^2{\left(\frac{dy}{dx}\right)}^2=y^2\left(b^2-a^2\right)\cos 2x$

$y^3\frac{d^{2}y}{d{x}^2}=y^2\left(b^2-a^2\right)\cos 2x-y^2{\left(\frac{dy}{dx}\right)}^2$

Add $y^4$ in the above equation.

$y^4+y^3\frac{d^{2}y}{d{x}^2}=y^4+y^2\left(b^2-a^2\right)\cos 2x-y^2{\left(\frac{dy}{dx}\right)}^2$

Substitute the values in the above equation.

$y^4+y^3\frac{d^{2}y}{d{x}^2}=y^4+y^2\left(b^2-a^2\right)\cos 2x-y^2{\left(\frac{dy}{dx}\right)}^2$

$=\left\{\begin{array}{l}\left(a^2{\cos }^{2}x+b^2{\sin }^{2}x\right)\left(b^2-a^2\right)\left({\cos }^{2}x-{\sin }^{2}x\right)-\\ {\left(\left(b^2-a^2\right)\sin x\cos x\right)}^2+{\left(a^2{\cos }^{2}x+b^2{\sin }^{2}x\right)}^2\end{array}\right\}$

$=a^2{b}^2{\cos }^{4}x+a^2{b}^2{\sin }^{2}x+2a^2{b}^2{\sin }^{2}x{\cos }^{2}x$

$=a^2{b}^2{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2$

$=a^2{b}^2$

Or it can be written as,

$\frac{d^{2}y}{d{x}^2}+y=\frac{1}{y}{\left(\frac{ab}{y}\right)}^2$