The solution of the differential equation $$(x^2 + y^2)dx - 5xy \, dy = 0$$, $$y(1) = 0$$, is :

$$(x^{2}+y^{2})\,dx-5xy\,dy=0 \quad -(1)$$

$$(x^{2}+y^{2})-5xy\frac{dy}{dx}=0$$

$$\frac{dy}{dx}=\frac{x^{2}+y^{2}}{5xy} =\frac{1}{5}\left(\frac{x}{y}+\frac{y}{x}\right) \quad -(2)$$

The right-hand side is a function of $$y/x$$, so set $$y=vx \;(\Rightarrow v=\tfrac{y}{x})$$.
Then $$\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}$$.

Substitute in $$(2)$$:

$$v+x\frac{dv}{dx}=\frac{1}{5v}\left(1+v^{2}\right)$$

$$x\frac{dv}{dx}=\frac{1}{5v}\left(1+v^{2}\right)-v =\frac{1+v^{2}-5v^{2}}{5v} =\frac{1-4v^{2}}{5v} \quad -(3)$$

$$\frac{5v}{1-4v^{2}}\,dv=\frac{dx}{x}$$

Integrate both sides.
Use the substitution $$w=1-4v^{2}\;\;(dw=-8v\,dv)$$ on the left:

$$\int\frac{5v}{1-4v^{2}}\,dv =-\frac{5}{8}\int\frac{dw}{w} =-\frac{5}{8}\ln|w| =-\frac{5}{8}\ln|1-4v^{2}| \quad -(4)$$

The right integral gives $$\int\frac{dx}{x}=\ln|x|$$.

$$-\frac{5}{8}\ln|1-4v^{2}|=\ln|x|+C \quad -(5)$$

$$\ln|1-4v^{2}|=-\frac{8}{5}\ln|x|+C_{1}$$

$$|1-4v^{2}|=K\,x^{-8/5},\qquad K=e^{C_{1}}\gt 0 \quad -(6)$$

Replace $$v=\dfrac{y}{x}$$:

$$\left|1-4\frac{y^{2}}{x^{2}}\right|=K\,x^{-8/5}$$

$$|x^{2}-4y^{2}|=K\,x^{2/5} \quad -(7)$$

Apply the initial condition $$y(1)=0$$.
From $$(7)$$ with $$x=1,\;y=0$$:

$$|1^{2}-4\cdot0^{2}|=K\cdot1^{2/5}\;\Longrightarrow\;1=K$$

$$K=1$$ and $$(7)$$ becomes

$$|x^{2}-4y^{2}|=x^{2/5}$$

$$|x^{2}-4y^{2}|^{5}=x^{2}$$