The solution of the differential equation $x \frac{d y}{d x}+y=x^3 y^6$, is

$2 x^7=5 y^5+C x^2 y^5$

The given differential equation can be written as

$\frac{1}{y^6} \frac{d y}{d x}+\frac{1}{x y^5}=x^2$

Let $y^{-5}=v$. Then,

$-5 y^{-6} \frac{d y}{d x}=\frac{d v}{d x} \Rightarrow y^{-6} \frac{d y}{d x}=-\frac{1}{5} \frac{d v}{d x}$

Substituting these values in the given differential equation, we get

$-\frac{1}{5} \frac{d v}{d x}+\frac{1}{x} v=x^2$

$\Rightarrow \frac{d v}{d x}-\frac{5}{x} v=-5 x^2$            .......(i)

This is the standard form of the linear differential equation having integrating factor

Integrating factor = $e^{\int-\frac{5}{x} d x}=e^{-5 \log x}=\frac{1}{x^5}$

Multiplying both sides of (i) by integrating factor and integrating w.r.t. $x$, we get

$v \frac{1}{x^5}=\int-5 x^2 . \frac{1}{x^5} d x$

$\Rightarrow \frac{v}{x^5}=\frac{5}{2} x^{-2}+C$

$\Rightarrow y^{-5} x^5=\frac{5}{2} x^{-2}+C$, which is the required solution.