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Prove It $\displaystyle \displaystyle \begin{align*} \left(x + y\right)\frac{dy}{dx} &= \frac{y^2}{x + 4y} \\ \frac{dy}{dx} &= \frac{y^2}{(x + y)(x + 4y)} \\ \frac{dy}{dx} &= \frac{y^2}{x^2 + 5x\,y + 4y^2} \\ \frac{dy}{dx} &= \frac{\left(\frac{y}{x}\right)^2}{1 + 5\left(\frac{y}{x}\right) + 4\left(\frac{y}{x}\right)^2} \end{align*}$
Let $\displaystyle \displaystyle \begin{align*} u = \frac{y}{x} \implies y = u\,x \implies \frac{dy}{dx} = u + x\,\frac{du}{dx} \end{align*}$ and the DE becomes
$\displaystyle \displaystyle \begin{align*} u + x\,\frac{du}{dx} &= \frac{u^2}{1 + 5u + 4u^2} \\ x\,\frac{du}{dx} &= \frac{u^2}{1 + 5u + 4u^2} - u \\ x\,\frac{du}{dx} &= \frac{u^2}{1 + 5u + 4u^2} - \frac{u\left(1 + 5u + 4u^2\right)}{1 + 5u + 4u^2} \\ x\,\frac{du}{dx} &= \frac{u^2 - u\left(1 + 5u + 4u^2\right)}{1 + 5u + 4u^2} \\ x\,\frac{du}{dx} &= \frac{u^2 - u - 5u^2 - 4u^3}{1 + 5u + 4u^2} \\ x\,\frac{du}{dx} &= -\frac{4u^3 + 4u^2 + u}{4u^2 + 5u + 1} \\ \left(\frac{4u^2 + 5u + 1}{4u^3 + 4u^2 + u}\right)\frac{du}{dx} &= -\frac{1}{x} \end{align*}$
You should be able to solve this now.