Let's get some formulas straight in the first place. You have
$$F(z) = \phi + i\psi
$$
Also, it is given that
$$u = \frac{\partial \phi}{\partial x}, \ \ v = \frac{\partial \phi}{\partial y}
$$
And no, what you have instead is this:
$$u = \frac{\partial \psi}{\partial y}, \ \ v = -\frac{\partial \psi}{\partial x}
$$
Differentiate in the $x$-direction or in the $y$-direction - it doesn't matter (why?):
$$ \frac{dF}{dz} = \frac{\partial F}{\partial x}\frac{\partial x}{\partial z}
\qquad ; \qquad
\frac{dF}{dz} = \frac{\partial F}{\partial y}\frac{\partial y}{\partial z}
$$
With
$$ \frac{\partial z}{\partial x} = 1 \quad \Longrightarrow \quad \frac{\partial x}{\partial z} = 1/1 = 1 \qquad ; \qquad
\frac{\partial z}{\partial y} = i \quad \Longrightarrow \quad \frac{\partial y}{\partial z} = 1/i = -i
$$
Hence in the $x$- or $y$-direction, take your favorite:
$$
\frac{dF}{dz} = 1 \left(\frac{\partial \phi}{\partial x} + i \frac{\partial \psi}{\partial x} \right) =
-i \left( \frac{\partial \phi}{\partial y} + i \frac{\partial \psi}{\partial y}\right)
= u - i v
$$
More about this subject in Cauchy-Riemann Equations (PDF).