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The given differential equation i.e., $\left(x^{2}+x y\right) d y=\left(x^{2}+y^{2}\right) d x$ can be written as:

$\frac{d y}{d x}=\frac{x^{2}+y^{2}}{x^{2}+x y}$               ...(1)

Let $F(x, y)=\frac{x^{2}+y^{2}}{x^{2}+x y}$

Now, $F(\lambda x, \lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{(\lambda x)^{2}+(\lambda x)(\lambda y)}=\frac{x^{2}+y^{2}}{x^{2}+x y}=\lambda^{0} \cdot F(x, y)$

This shows that equation (1) is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

$\frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $v$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x^{2}+(v x)^{2}}{x^{2}+x(v x)}$

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

$\Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{1+v}-v=\frac{\left(1+v^{2}\right)-v(1+v)}{1+v}$

$\Rightarrow x \frac{d v}{d x}=\frac{1-v}{1+v}$

$\Rightarrow\left(\frac{1+v}{1-v}\right)=d v=\frac{d x}{x}$

$\Rightarrow\left(\frac{2-1+v}{1-v}\right) d v=\frac{d x}{x}$

$\Rightarrow\left(\frac{2}{1-v}-1\right) d v=\frac{d x}{x}$

Integrating both sides, we get:

$-2 \log (1-v)-v=\log x-\log k$

$\Rightarrow v=-2 \log (1-v)-\log x+\log k$

$\Rightarrow v=\log \left[\frac{k}{x(1-v)^{2}}\right]$

$\Rightarrow \frac{y}{x}=\log \left[\frac{k}{x\left(1-\frac{y}{x}\right)^{2}}\right]$

$\Rightarrow \frac{y}{x}=\log \left[\frac{k x}{(x-y)^{2}}\right]$

$\Rightarrow \frac{k x}{(x-y)^{2}}=e^{\frac{y}{x}}$

$\Rightarrow(x-y)^{2}=k x e^{-\frac{y}{x}}$

This is the required solution of the given differential equation.