My solution assumes that $f$ is $C^1$.
Note that implicit in the problem description is the fact that $x,y$ are smooth (or sufficiently smooth, at least) functions of time, and that the skier's $y$ position can be written as a function of the $x$ coordinate.
(i) We are given that the boat at $(0,y)$ and the skier at $(x,f(x))$ are a distance $L$ apart, this gives the constraint $x^2+(y-f(x))^2 = L^2$. The boat travels 'north' (presumably in a straight line, although the question did not make this explicit) starting from the origin, hence we have $y \ge 0$. The skier starts at $(L,0)$, hence we have $f(L) = 0$.
We are also given that the skier points towards the boat. The skier's direction is presumably the tangent to movement, we have $\frac{d}{dx} (x,f(x)) = \lambda (-x,y-f(x))$, which gives $f'(x) = -\frac{1}{x}(y-f(x))$.
Substituting this into the constraint gives $x^2(1+(f'(x))^2) = L^2$, from which we get $f'(x) = 0$ iff $|x|=L$, in particular, $f'(x) $ does not change sign on $(0,L)$ (here I am relying on $f'$ being continuous).
Suppose $f'(x) >0$ on $(0,L)$, then since $f(L) = 0$, we have $f(x) < 0$ for $x \in (0,L)$, and since $y \ge 0$, this gives $f'(x) = -\frac{1}{x}(y-f(x)) < 0$, a contradiction. Hence $f'(x) < 0$ on $(0,L)$, and then we have $y > f(x)$.
Then we can solve the constraint for $y-f(x)$ to get $y-f(x) = +\sqrt{L^2-x^2}$, from which we obtain $f'(x) = -\frac{\sqrt{L^2-x^2}}{x}$, as required.
(ii) We have
$f'(x) = -\frac{\sqrt{L^2-x^2}}{x}$.
To reduce clutter, let $\phi(t) = f(Lt)$, then $\phi'(t) = Lf'(Lt) = -L \frac{\sqrt{1-t^2}}{t}$. Then we have $\phi(1)-\phi(t) = \int_t^L \phi'(\tau) d\tau$, or $\phi(t) = -\int_t^L \phi'(\tau) d\tau$, since a boundary condition is $\phi(1) = 0$. It is straightforward to compute $\int_t^1 \frac{\sqrt{1-\tau^2}}{\tau} d \tau = \frac{1}{2}\log \frac{1+\sqrt{1-t^2}}{1-\sqrt{1-t^2}} - \sqrt{1-t^2}$, from which we get $f(x) = \phi(\frac{x}{L})$, or
$$ f(x) = \frac{L}{2}\log \left( \frac{L+\sqrt{L^2-x^2}}{L-\sqrt{L^2-x^2}} \right) - \sqrt{L^2-x^2} $$
You can use a formula, although I think it's not too difficult to just go through the steps. I would draw a picture first:
You are given that $\vec{p} = (1,0,1)$ and you already found $\vec{m} = (1, -2, 4)$ and $\vec{l}_0 = (1,2,-1)$. Now it's a matter of writing an expression for $\vec{l}(t) - \vec{p}_0$:
\begin{align}
\vec{l}(t) - \vec{p}_0 =&\ (\ (t + 1) - 1\ ,\ (-2t + 2) - 0\ ,\ (4t - 1) - 1\ )\\
=&\ (\ t\ ,\ -2t + 2\ ,\ 4t - 2\ )
\end{align}
Now you dot this with the original slope of the line (recall that $\vec{l}(t) - \vec{p}_0$ is the slope of the line segment connecting the point and the line). When this dot product equals zero, you have found $t_0$ and thus $\vec{x}_0$:
\begin{align}
\vec{m} \circ (\vec{l}(t) - \vec{p}_0) =&\ (1,-2,4)\circ(\ t\ ,\ -2t + 2\ ,\ 4t - 2\ ) \\
=&\ t + 4t - 4 + 16t - 8 \\
=&\ 21t - 12
\end{align}
Setting this to $0$ gives that $21t_0 - 12 = 0 \rightarrow t_0 = \frac{4}{7}$. This gives the point $\vec{x}_0$ as:
\begin{align}
\vec{x}_0 =&\ \vec{l}(t_0) = (\ \frac{4}{7} + 1\ ,\ -\frac{8}{7} + 2\ ,\ \frac{16}{7} - 1\ ) \\
=&\ \frac{1}{7}(11, 6, 9)
\end{align}
So finally the distance would be the distance from $\vec{p}_0$ to $\vec{x}_0$:
\begin{align}
d =&\ \sqrt{\left(\frac{11}{7} - 1\right)^2 + \left(\frac{6}{7} - 0\right)^2 + \left(\frac{9}{7} - 1\right)^2}\\
=&\ \sqrt{\left(\frac{4}{7}\right)^2 + \left(\frac{6}{7}\right)^2 + \left(\frac{2}{7}\right)^2} \\
=&\ \frac{1}{7}\sqrt{4^2 + 6^2 + 2^2}\\
=&\ \frac{1}{7}\sqrt{56} \\
=&\ \frac{2}{7}\sqrt{14}
\end{align}
...or perhaps $\sqrt{\frac{8}{7}}$ is more appealing.
Extra Info
There's no need to worry about whether or not my 2D picture is really representative--it is. No matter how high the dimensions of the problem, the problem itself can always be mapped to exactly 2 dimensions unless the point is on the line--then it's a 1 dimensional problem--which of course we can represent in 2 dimensions just as we can represent this 2 dimensional problem in much higher ones.
Best Answer
Hint: $\;\vec{AP}=\dfrac{\vec{AP} \cdot \vec{AB}}{\left|\vec{AB}\right|^2} \, \vec{AB} + \vec{d}\,$, so $\vec{d} = \vec{AP} - \dfrac{\vec{AP} \cdot \vec{AB}}{\left|\vec{AB}\right|^2} \, \vec{AB}\,$ where the RHS is easily calculated.