Given equation is $\frac{1}{\sqrt{y}}\frac{dy}{dx}+\frac{x}{1-x^2}\sqrt{y}=x$
Let $2\sqrt{y}=\nu$
$\Rightarrow \frac{1}{\sqrt{y}}\frac{dy}{dx}=\frac{d\nu }{dx}$
Thus, we have $\frac{d\nu }{dx}+\frac{x}{2\left(1-x^2\right)}\cdot \nu =x$
$\therefore \mathrm{I}.\mathrm{F}.=e^{\int \frac{x}{2\left(1-x^2\right)}dx}$
$=e^{-\frac{1}{4}\ln \left(1-x^2\right)}$
$={\left(1-x^2\right)}^{-\frac{1}{4}}$
Thus, the solution is $\nu {\left(1-x^2\right)}^{-\frac{1}{4}}=\int x{\left(1-x^2\right)}^{-\frac{1}{4}}dx$
or $\nu \cdot {\left(1-x^2\right)}^{-\frac{1}{4}}=-\frac{2}{3}{\left(1-x^2\right)}^{\frac{3}{4}}+c^{"}$
or $2\sqrt{y}=-\frac{2}{3}\left(1-x^2\right)+c^{"}{\left(1-x^2\right)}^{\frac{1}{4}}$
$\Rightarrow \sqrt{y}=-\frac{\left(1-x^2\right)}{3}+c^{\text{'}}{\left(1-x^2\right)}^{\frac{1}{4}}$

$\Rightarrow \sqrt{9y}=-\left(1-x^2\right)+c{\left(1-x^2\right)}^{\frac{1}{4}}$
$\Rightarrow f\left(x\right)=1-x^2$
$\therefore f\left(x\right)$ is an even function