Q: Find all functions that are analytic and satisfy $(\operatorname{Re}(f(z)))^2 = \operatorname{Im}(f(z))$ I've been trying to find all functions $f(z):\Bbb{C}\longrightarrow\Bbb{C}$ that are analytic and satisfy
$$(\operatorname{Re}(f(z)))^2 = \operatorname{Im}(f(z))$$
After plugging in $f(z)=x+iy$, I got $f(z)=x+ix^2$. 
These kind of functions never satisfy Cauchy-Riemann equations, since $\mathcal u_x=1 \neq 0=v_y$. 
Does it follow from this argumentation that no such analytic $f(z)$ exists?

You might have rushed your analysis a little bit. The condition you have given is equivalent to saying that
$$ f(x+  iy) = u(x,y) + i u(x,y)^2$$
Computing the Cauchy-Riemann equations we have
$$ \frac{\partial u}{\partial x} = 2u\frac{\partial u}{\partial y}$$
and
$$ \frac{\partial u}{\partial y} = -2u\frac{\partial u}{\partial x}$$
Combining these two we have
$$\frac{\partial u}{\partial x} = - 4 u^2 \frac{\partial u}{\partial x} \iff \frac{\partial u}{\partial x} \left(1 + 4u^2\right) = 0$$
Since $1 + 4u^2 > 0$ for all $u \in \mathbb{R}$ we must have $\frac{\partial u}{\partial x} = 0$. In similar fashion we may also compute
$$ \frac{\partial u}{\partial y} = -  4u^2 \frac{\partial u}{\partial y} \iff\frac{\partial u}{\partial y} \left(1 + 4u^2\right) = 0 $$
From which we may conclude that $\frac{\partial u}{\partial x} = 0$. So $u(x,y)$ must be a constant function. Since we did not specify any specific value of the constant function $u(x,y)$ then it follows that it works for all values. Thus functions with a constant real part which satisfy the hypothesis you have given are all solutions.
