Find a curve $C$ in the first quadrant in $\mathbb{R}^2$ passing through $(3,2)$ with the property that if $P = (x_0,y_0)$ lies on $C$, then $P$ is the midpoint of the tangent line to $C$ at $P$ contained in the first quadrant.
I started by realizing $(3,2)$ must be the midpoint of a tangent line, so it must be the midpoint of the line going through $(0,4)$ and $(4,0)$, i.e. at point $(3,2)$ the curve $C$ has slope $\frac{-2}{3}$. I'm just not sure how to generalize this to all points on $C$.
Unfortunately, you seem to be off track: $(3,2)$ is not the midpoint of the segment from $(4,0)$ to $(0,4)$--indeed, it doesn't even lie on said segment--though given your slope, I suspect these points may have been mistyped.
One thing we can be sure of is this: regardless of our choice of $(x_0,y_0)$ in the first quadrant (on the curve), the tangent line at that point must intersect the positive $x$- and $y$-axes, or the midpoint condition is impossible. Consequently, the slope of the tangent at such a point is necessarily negative. Call this slope $m(x_0,y_0),$ so that the tangent line is given by $$y-y_0=m(x_0,y_0)(x-x_0).$$ Find the $x$- and $y$-intercepts of this line, find their midpoint, and set the midpoint equal to $(x_0,y_0).$ Can you take it from there?