if $f(x)$ such $x^2\cdot f'(x)+2xf(x)=\frac{e^x}{x}$ then $f'(x)\ge 0$ 
Let function $f(x)$ such $$x^2\cdot f'(x)+2xf(x)=\dfrac{e^x}{x},f(2)=\dfrac{e^2}{8}$$
  show that
  $$f'(x)\ge 0,\forall x>0$$

I thought
$4f'(2)+4f(2)=\dfrac{e^2}{2}$,then $f'(2)=0$,also
$$2xf'(x)+x^2f''(x)+2f(x)+2xf'(x)=\dfrac{e^x\cdot x-e^x}{x^2}\Longrightarrow f''(2)=0$$Well and now I'm stuck and don't know how to proceed

We have
\begin{eqnarray*}
x^{2}\partial _{x}f(x)+2xf(x) &=&\partial _{x}\{x^{2}f(x)\}=\frac{\exp [x]}{x
} \\
x^{2}f(x) &=&4f(2)+\int_{2}^{x}dy\frac{\exp [y]}{y}=\frac{e^{2}}{2}%
+\int_{2}^{x}dy\frac{\exp [y]}{y} \\
f(x) &=&\frac{1}{x^{2}}\{\frac{e^{2}}{2}+\int_{2}^{x}dy\frac{\exp [y]}{y}\}
\\
\partial _{x}f(x) &=&-\frac{2}{x^{3}}\{\frac{e^{2}}{2}+\int_{2}^{x}dy\frac{
\exp [y]}{y}\}+\frac{1}{x^{2}}\frac{\exp [x]}{x}=\frac{1}{x^{3}}\{\exp
[x]-e^{2}-2\int_{2}^{x}dy\frac{\exp [y]}{y}\}\\=\frac{1}{x^{3}}g(x)
\end{eqnarray*}
For $x<2$ we have
\begin{eqnarray*}
g(x) &=&\exp [x]-e^{2}-2\int_{2}^{x}dy\frac{\exp [y]}{y}=\exp
[x]-e^{2}+2\int_{x}^{2}dy\frac{\exp [y]}{y} \\
&\geqslant &\exp [x]-e^{2}+2\int_{x}^{2}dy\frac{\exp [y]}{2}=\exp
[x]-e^{2}+\int_{x}^{2}dy\exp [y]=0
\end{eqnarray*}
so $\partial _{x}f(x)\geqslant 0$.
For $x>2$
$$
g(x)=\partial _{x}\{\exp [x]-e^{2}-2\int_{2}^{x}dy\frac{\exp [y]}{y}\}=\exp
[x]-2\frac{\exp [x]}{x}=(1-\frac{2}{x})\exp [x]>0
$$
so $g(2)=0$ and $g(x)$ is an increasing function for $x>2$ and we conclude
that $\partial _{x}f(x)>0$ for $x>2$.
