Solve the following homogeneous differential equation Initial value problem: $\displaystyle \frac{dy}{dx}= \frac{y}{x}+2 \frac{x^2}{y^2}$, $y(1)=1$.  Can anyone help

Again, start with $y = ux$ so that $\frac{dy}{dx} = u + x\frac{du}{dx}$ (applying the simple chain rule).
Substituting this into the original equation, along with
$ \frac{y}{x} = u $ and $ {\frac{x}{y}}^2 = u^{-2} $ results in 
$u + x\frac{du}{dx} = u + 2u^{-2}$.
Since the u terms cancel on each side, the equation may be rearranged as 
$u^2du = 2x^{-1} dx$.
At $x=x_0 = 1$, $y=y_0=1$ and $u=u_0 = 1$, so we can integrate the left side from 1 to $u$ and integrate the right side from 1 to $x$. This yields $\frac{1}{3}\left(u^3-1\right)=2ln(x)$ -> $u^3=1+6ln(x)$  -> $u={\left(1+6 ln x\right)}^\frac{1}{3}$.
Since u = $\frac{y}{x}$, the final result is $y=x{\left(1+6lnx\right)}^\frac{1}{3}$.
No answer is complete without validating, so let's check the terms:
$$\frac{dy}{dx} = {\left(1+6lnx\right)}^\frac{1}{3} + x\frac{1}{3}{\left(1+6lnx\right)}^\frac{-2}{3}\left(6\right)\frac{1}{x} =  {\left(1+6lnx\right)}^\frac{1}{3} + 2 {\left(1+6lnx\right)}^\frac{-2}{3}
$$
$$ \frac{y}{x} = {\left(1+6lnx\right)}^\frac{1}{3}$$
$$ 2\frac{x^2}{y^2} = 2{\left(1+6lnx\right)}^\frac{-2}{3} $$
And at $x = 1$, $y = 1{\left(1+6ln (1)\right)}^\frac{1}{3} = 1$
