Finding the equation of a function 
Find the function $y=f(x)$ whose graph is the curve $C$ passing through the point $(2,1)$ and satisfying the following property: each point $(x,y)$ of $C$ is the midpoint of $L(x,y)$ where $L(x,y)$ denotes the segment of the tangent line to $C$ at $(x,y)$ which lies in the first quadrant.

Okay, so with $(2,1)$ being the midpoint of the line segment $L(2,1)$, we have that the $L(2,1)$ must lie on the line $y=2-x/2$. This gives us that that slope of $C$ at $(2,1)$ is $-\frac{1}{2}$. Now I'm not really sure where to go from here.

At any point of your curve you have information about the slope of the curve at your point. This means that you have to solve a differential equation.
Suppose $(x_0,y_0) \in C$ is a point of your curve. Then (I skip boring computations) you have that the slope of the curve at $x_0$ is 
$$f'(x_0)=-\frac{y_0}{x_0}$$
so you have to solve the Cauchy problem
$$\left\{ \begin{matrix}
y'=-\frac{y}{x} \\
y(2)=1
\end{matrix} \right.$$
with constraints $y >0, x>0$.
The solution is $f(x)=\frac{2}{x}$
EDIT: How did I find $f'(x_0)=-\frac{y_0}{x_0}$? Here I show my boring (actually there is nothing smart in here) computations.
Let $m= f'(x_0)$. The line $L$ passing through $(x_0,y_0)$ of slope $m$ has equation
$$L: \ \ y-y_0=m(x-x_0)$$
Intersecting $L$ with the $x$ axis we get the point $(-\frac{y_0}{m}+x_0, 0)$.
Intersecting $L$ with the $y$ axis we get the point $(0, y_0-mx_0)$.
So we get the conditions
$$\left\{\begin{matrix}2x_0 = -\frac{y_0}{m}+x_0 \\
2y_0 = y_0-mx_0\end{matrix}\right.$$
which are equivalent to $m= -\frac{y_0}{x_0}$.
