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Hello and welcome to MHB, talha!

First, I need to say that our goal here at MHB is to help people solve their problems, not solve it for them. You will learn more through guided effort than simply being given the solution.

Second, I suspect the exponents on $e$ are squared as follows:

\(\displaystyle y'=\frac{2xye^{\left(\frac{x}{y} \right)^2}}{y^2+y^2e^{\left(\frac{x}{y} \right)^2}+2x^2e^{\left(\frac{x}{y} \right)^2}}\)

Is this correct?

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This is how I would work the problem:

\(\displaystyle y'=\frac{2xye^{\left(\frac{x}{y} \right)^2}}{y^2+y^2e^{\left(\frac{x}{y} \right)^2}+2x^2e^{\left(\frac{x}{y} \right)^2}}\)

We observe that we must have $y\ne0$ so multiplying the right by \(\displaystyle 1=\frac{y^{-2}}{y^{-2}}\) will not result in the loss of any trivial solution:

\(\displaystyle y'=\frac{2\dfrac{x}{y}e^{\left(\frac{x}{y} \right)^2}}{1+e^{\left(\frac{x}{y} \right)^2}+2\left(\dfrac{x}{y} \right)^2e^{\left(\frac{x}{y} \right)^2}}\)

Now the ODE is in the form \(\displaystyle y'=f\left(\frac{x}{y} \right)\) and we may use the substitution:

\(\displaystyle v=\frac{y}{x}\implies y=vx\,\therefore y'=v+xv'\)

Hence, the ODE may now be written:

\(\displaystyle v+xv'=\frac{\dfrac{2}{v}e^{v^{-2}}}{1+e^{v^{-2}}+\dfrac{2}{v^2}e^{v^{-2}}}\)

Multiplying the right side by \(\displaystyle 1=\frac{v^2}{v^2}\) we obtain:

\(\displaystyle v+xv'=\frac{2ve^{v^{-2}}}{v^2+v^2e^{v^{-2}}+2e^{v^{-2}}}\)

Subtracting through by $v$ and simplifying, we get:

\(\displaystyle xv'=-\frac{v^3\left(1+e^{v^{-2}} \right)}{v^2\left(1+e^{v^{-2}} \right)+2e^{v^{-2}}}\)

Separation of variables yields:

\(\displaystyle \frac{v^2\left(1+e^{v^{-2}} \right)+2e^{v^{-2}}}{v^3\left(1+e^{v^{-2}} \right)}\,dv=-\frac{1}{x}\,dx\)

On the left side, dividing each term by \(\displaystyle 1+e^{v^{-2}}\) allows us to write:

\(\displaystyle \left(\frac{1}{v}-\frac{-2v^{-3}e^{v^{-2}}}{1+e^{v^{-2}}} \right)\,dv=-\frac{1}{x}\,dx\)

Now observing that:

\(\displaystyle \frac{d}{dv}\left(1+e^{v^{-2}} \right)=-2v^{-3}e^{v^{-2}}\)

we may integrate as follows:

\(\displaystyle \ln|v|-\ln\left(1+e^{v^{-2}} \right)=-\ln|x|+C\)

Back-substituting for $v$, we obtain:

\(\displaystyle \ln\left|\frac{y}{x} \right|-\ln\left(1+e^{\left(\frac{x}{y} \right)^2} \right)=-\ln|x|+C\)

Adding \(\displaystyle \ln|x|\) to both sides, we obtain:

\(\displaystyle \ln|y|-\ln\left(1+e^{\left(\frac{x}{y} \right)^2} \right)=C\)

Using a property of logs, we may rewrite the left side as:

\(\displaystyle \ln\left|\frac{y}{1+e^{\left(\frac{x}{y} \right)^2}} \right|=C\)

Converting from logarithmic to exponential form, we find:

\(\displaystyle \left|\frac{y}{1+e^{\left(\frac{x}{y} \right)^2}} \right|=e^{C}\)

If we redefine the constant on the right as any positive value (since $y\ne0$), we have:

\(\displaystyle \left|\frac{y}{1+e^{\left(\frac{x}{y} \right)^2}} \right|=C\) where \(\displaystyle 0<C\)

Now, if we redefine the constant again as any real value except zero, we may write:

\(\displaystyle \frac{y}{1+e^{\left(\frac{x}{y} \right)^2}}=C\) where \(\displaystyle C\ne0\)

And if we so choose, we may arrange this as the implicit solution:

\(\displaystyle y=C\left(1+e^{\left(\frac{x}{y} \right)^2} \right)\)