= Solution
Set
$$
F(\eta)=f'(\eta)
=\frac{5-\cosh z}{1+\cosh z},
\qquad
z=\sqrt2\eta+c.
$$
Writing $C=\cosh z$, direct differentiation gives
$$
F''(\eta)=\frac{12(C-2)}{(1+C)^2}.
$$
On the other hand,
$$
1-F^2
=\frac{(1+C)^2-(5-C)^2}{(1+C)^2}
=\frac{12(C-2)}{(1+C)^2}.
$$
Thus $F''=1-F^2$, which is precisely $f'''=1-(f')^2$. The far-field condition is automatic because $F\to-1$. No slip requires
$$
0=F(0)=\frac{5-\cosh c}{1+\cosh c},
$$
so
$$
\boxed{c=\pm\operatorname{arcosh}5}.
$$
For $c=+\operatorname{arcosh}5$, $F$ decreases directly from $0$ to $-1$, matching flow everywhere toward the sink. For the negative root, $F$ initially becomes positive and creates a reverse-flow region next to the wall. The physically likely choice is therefore
$$
\boxed{c=+\operatorname{arcosh}5}.
$$
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