Solve $\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$ for $y(1) = 2$
I began by simplifying such that: $\frac{dy}{dx} = \frac{x}{y} - \frac {y}{x}$
Then, I set $v = \frac {y}{x}$
Therefore: $y = xv$ and $y'$ = $v$ + $xv'$
Substituting back: $v + xv' = \frac{x}{xv} - \frac{xv}{x}$
or $v + xv' = \frac{1}{v} - v$
To continue, should $x$ be divided from both sides to allow for use of an integrating factor?
To continue with your method then
( You have a sign mistake ) $$v + xv' = 1/v \color{red}{+} v$$ $$v'vx=1$$
It's separable $$\int vdv=\int \frac {dx}x$$ $$v^2=2 \ln |x|+C$$ Substitute back $v=\frac y x$
Another approach
$$y'xy=x^2+y^2$$ Note that $$(y^2)'=2yy'$$ $$\frac x2(y^2)'=x^2+y^2$$ Substitute $y^2=z$ $$\frac x2z'=x^2+z$$ It's a first order differential equation $$z'-\frac 2xz=2x$$ Multiply by $x^2$ $$z'x^2-2xz=2x^3$$ Divide by $x^4$ $$\left(\frac z {x^2}\right )'=\frac 2x$$ Integrate $$\left(\frac z {x^2}\right )=2 \ln |x|+K$$ $$z=2 x^2\ln |x|+Kx^2$$ $$z(1)=y^2(1)=4 \implies K=z=4$$ $$\boxed {y(x)=\pm |x|\sqrt {2\ln |x|+4}}$$