Given:

$\left(y+1\right){\tan }^{2}xdx+\tan xdy+ydx=0$

$\Rightarrow \left[\left(y+1\right){\tan }^{2}x+y\right]dx+\tan xdy=0$

$\Rightarrow \tan x\left(\frac{dy}{dx}\right)+\left(y+1\right){\tan }^{2}x+y=0$

$\Rightarrow \frac{dy}{dx}+\left(1+y\right)\tan x=-y\cot x$

$\Rightarrow \frac{dy}{dx}+y\left(\tan x+\cot x\right)=-\tan x$

This is a linear differential equation of the form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$.

$\text{ I.F }=e^{\int (\tan x+\cot x)dx}$

$=e^{\int \left(\frac{{\tan }^{2}x+1}{\tan x}\right)dx}$

$=e^{\int \left(\frac{{\sec }^{2}x}{\tan x}\right)dx}$

$=e^{{\log }_e\left|\tan x\right|}$

$=\tan x \left(\because x\in \left(0,\frac{\pi }{2}\right)\right)$

Solution is

$y\tan x=\int -{\tan }^{2}xdx+c$

$\Rightarrow y\tan x=\int \left(1-{\sec }^{2}x\right)dx+c$

$\Rightarrow y\tan x=x-\tan x+c$

Now,

$\underset{x\to 0^{+}}{\lim }xy=1$

$\Rightarrow \underset{x\to 0^{+}}{\lim }\left(\frac{x}{\tan x}\right)\left(x-\tan x+c\right)=1$

$\Rightarrow 1(0-0+c)=1\Rightarrow c=1$

Then the function is

$y\tan x=x-\tan x+1$

Put $x=\frac{\pi }{4}$,

$y\left(\frac{\pi }{4}\right)\tan \left(\frac{\pi }{4}\right)=\left(\frac{\pi }{4}\right)-\tan \left(\frac{\pi }{4}\right)+1$

$\Rightarrow y\left(\frac{\pi }{4}\right)=\frac{\pi }{4}$