Solve this differential equation $\Biggr(\frac{dy}{dx} -1\Bigg)\Bigg(y-x\frac{dy}{dx}\Bigg)=\frac{dy}{dx}$

$\Biggr(\frac{dy}{dx} -1\Bigg)\Bigg(y-x\frac{dy}{dx}\Bigg)=\frac{dy}{dx}\implies(p-1)(y-xp)=p$, where $p\equiv \frac{dy}{dx}$

$\implies y-xp=\frac{p}{p-1}=\implies y=xp+\frac{p}{p-1}$ . . . . $(1)$

which is known as Clairaut’s equation.

So the general solution of the differential equation $(1)$ is

$y=cx+\frac{c}{c-1}$, where $c$ is the integrating constant.

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If you once show that a differential equation is of Clairaut’s form i.e., of the form $$y=px+f(p)$$

Then for general solution of this kind of ODE, just replace $p\equiv \frac{dy}{dx}$ by the integrating constant $c$. i.e., its general solution is $$y=cx+f(c)$$

For more details about the solution of Clairaut’s equation, you may follow the reference: "http://www.library.gscgandhinagar.in/assets/admin/images/MAT-102(UNIT1,2).pdf"

We have $y(x)=px+\frac{p}{-1+p}$. Where $p=\frac{dy}{dx}$ This is the Clairaut's equation . The solution will he of the form $y=cx+f(c)$ .Differentiating we have $dy=pdx+xdp+\frac{dp}{(1-p)^2}$ we can write $dy=pdx$ the equation becomes $-xdp=\frac{dp}{(1-p)^2}$ assuming $dp\neq 0$ we have $x=\frac{1}{(1-p)^2}$ put this value in the first equation and get an expression for $y(p)$. The $x(p),y(p) $ are called singular solutions.

Since nmasanta already posted the answer I don't edit my calculus.

I only add the singular solution $y=x+1\pm 2\sqrt{x}$ and a graph of the complete solutions of the ODE :