Given:$\frac{y}{x}\frac{dy}{dx}=\left[\frac{y^2}{x^2}+\frac{\varphi \left(\frac{y^2}{x^2}\right)}{{\varphi }^{\text{'}}\left(\frac{y^2}{x^2}\right)}\right] ....\left(1\right)$ 

Let $\frac{y}{x}=t$

$\Rightarrow y=xt$

$\Rightarrow \frac{dy}{dx}=t+x·\frac{dt}{dx}$

$\therefore t\left(t+x\frac{dt}{dx}\right)=\left(t^2+\frac{\varphi \left(t^2\right)}{{\varphi }^{\text{'}}\left(t^2\right)}\right)$

$\Rightarrow xt\frac{dt}{dx}=\frac{\varphi \left(t^2\right)}{\varphi \text{'}\left(t^2\right)}$

$\Rightarrow \frac{t·{\varphi }^{\text{'}}\left(t^2\right)}{\varphi \left(t^2\right)}dt=\frac{1}{x}dx$
Integrating both sides

$\int \frac{t·{\varphi }^{\text{'}}\left(t^2\right)}{\varphi \left(t^2\right)}dt=\int \frac{1}{x}dx$

Let $\varphi \left(t^2\right)=p$

$\Rightarrow {\varphi }^{\text{'}}\left(t^2\right).2t=dp$

$\Rightarrow \frac{1}{2}\int \frac{1}{p}dp=\int \frac{1}{x}dx$

$\Rightarrow \frac{1}{2}\ln p=\ln x+C$

$\Rightarrow \frac{1}{2}\ln \varphi \left(t^2\right)=\ln x+C$

$\Rightarrow \frac{1}{2}\ln \left(\varphi \left(\frac{y^2}{x^2}\right)\right)=\ln x+C ...\left(2\right)$

If $x=1, y=-1$ then $\mathrm{C}=\frac{1}{2}\ln \left(\varphi \left(1\right)\right)$

Substituting value of $C$ in $\left(2\right)$

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

$\Rightarrow$$\ln \left(\varphi \left(\frac{y^2}{x^2}\right)\right)=\ln x^2+\ln \left(\varphi \left(1\right)\right)$

If $x=2$ then 

$\ln \left(\varphi \left(\frac{y^2}{4}\right)\right)=\ln 4+\ln \left(\varphi \left(1\right)\right)$

SO, $\varphi \left(\frac{y^2}{4}\right)=4\varphi \left(1\right)$