= Solution
Put $p=y'$ and differentiate the general form of <Clairaut's equation>:
$$
p=p+xp'+f'(p)p'.
$$
Cancellation gives
$$
\boxed{[x+f'(p)]\frac{dp}{dx}=0}.
$$
For $f(p)=1/(p-1)$ the equation is defined only when $p\ne1$. There are two types of smooth branch.
First, constant $p=C\ne1$ gives the straight-line family
$$
\boxed{y=Cx+\frac1{C-1},\qquad C\in\mathbb R\setminus\{1\}}.
$$
Second, $x+f'(p)=0$ gives $x=1/(p-1)^2$, so $x>0$ and $p=1\pm x^{-1/2}$. Substituting in the original equation gives its two singular <envelope> branches,
$$
\boxed{y=x\pm2\sqrt x\qquad(x>0)}.
$$
Their <derivatives> are indeed $1\pm1/\sqrt x$, verifying the required value of $p$. There is no finite-slope continuation of either <envelope> through $x=0$, and no real <envelope> branch for $x<0$.
For completeness, if “solution” requires only a <continuous> first <derivative>, one can join a line from the family to the corresponding <envelope> at their tangency point $x=(C-1)^{-2}$. The <function> and its first <derivative> match there, so such piecewise branches also solve the original first-order equation. They are not twice continuously <differentiable> at a nontrivial join. The lines and <envelope> branches above exhaust the smooth branches; allowing these tangent joins supplies the additional $C^1$ solutions.
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