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
$$ f(x) = \frac{L}{2}\log \left( \frac{L+\sqrt{L^2-x^2}}{L-\sqrt{L^2-x^2}} \right) - \sqrt{L^2-x^2} $$
If you have heard contour map, which may appear in your geography class. Think the solid as a mountain and draw the $\color{red}{contour\ line}$ first.
Steps to draw contour line(in a two dimension axes):
$$z=f(x,y)=4-x-2y$$ let $z=0$ then draw $$0=4-x-2y$$ let$z=0.5$ then
draw $$0.5=4-x-2y$$ let$z=1$ then draw $$1=4-x-2y$$
Once you get a contour line graph then you can imagine to draw a solid, which seems not so hard to draw in this case.
Finally, if you want to check your graph, go to http://www.wolframalpha.com/input/?i=z%3D4-x-2y
Best Answer
To give you a sense of what to expect in Calc III, Paul Dawkins of Lamar University has lecture notes online: http://tutorial.math.lamar.edu/Classes/CalcIII/CalcIII.aspx He also has Calc I and Calc II notes that might help to direct your self study.
MIT OpenCourseware has a set of video lectures and course materials on Multivariable Calculus along with other Calculus courses: http://ocw.mit.edu/courses/mathematics/