The Weierstrass-Enneper representation, the Gauss map Lemma: Let $x:S\to\mathbb{R}^3$ be a conformal minimal immersion of a Riemann surface. The 1-forms $f_k=(x_{k,u}-ix_{k,v})dz$ satisfy: $$ \sum_kf_k^2=0\qquad (1)\qquad \&\qquad \sum_k|f_k|^2\not=0\qquad (2)$$
Conversely, given a Riemann surface and holomorphic 1-forms $f=(f_1,f_2,f_3)$, satisfying (1), (2) and the period condition $$\Re\int_c f=0$$ for any rectifiable Jordan curve in $S$, then the map $x:S\to\mathbb{R}^3$, defined by $$x(p)=\Re\int_{p_0}^p f$$ is a conformal minimal immersion of $S$.
Problem: Define $g=\frac{f_3}{f_1-if_2}$.
Let $St(x,y,z)=\frac{x+iy}{1-z}$ be the stereographic projection through the north pole and onto the equatorial plane, and $N$ be the Gauss map induced by $x$. If we can show that $g=St\circ N$, the Weierstrass-Enneper representation theorem follows easily from the lemma.
Attempt: Two approaches are possible. One is to show the equality directly. Another is to verify the identity $$N=\left(\frac{2 \mathrm{Re}(g)}{1 + \bar g g}, \frac{2 \mathrm{Im}(g)}{1 + \bar g g}, \frac{1 - \bar g g}{1 + \bar g g}\right)$$
We will follow the first approach.
$$x_{,u}\times x_{,v}=(x_{2,u}x_{3,v}-x_{3,u}x_{2,v},\quad x_{3,u}x_{1,v}-x_{1,u}x_{3,v},\quad x_{1,u}x_{2,v}-x_{2,u}x_{1,v})$$
$$\& \qquad x_{2,u}x_{3,v}-x_{3,u}x_{2,v}=\Im(x_{2,u}x_{3,u}-x_{3,v}x_{2,v}+i(x_{2,u}x_{3,v}-x_{3,u}x_{2,v}))=\Im f_2\bar{f_3}$$
$$\& \qquad x_{3,u}x_{1,v}-x_{1,u}x_{3,v}=\Im f_3\bar{f_1}$$
$$\& \qquad x_{1,u}x_{2,v}-x_{2,u}x_{1,v}=\Im f_1\bar{f_2}$$
$$\Rightarrow x_{,u}\times x_{,v}=\Im(f_2\bar{f_3},f_3\bar{f_1},f^1\bar{f_2})=\frac{1}{2} f\times \bar{f}$$
We will verify only the first component of the last equality. The others follow analogously. $$(f\times\bar{f})_1=f_2\bar{f_3}-f_3\bar{f_2}=(x_{2,u}-ix_{2,v})(x_{3,u}+ix_{3,v})=$$
$$x_{2,u}x_{3,u}+x_{2,v}x_{3,v}-x_{2,u}x_{3,u}-x_{2,v}x_{3,v}+i(-x_{2,v}x_{3,u}+x_{2,u}x_{3,v}-x_{2,v}x_{3,u}+x_{2,u}x_{3,v})=2\Im f_2\bar{f_3}$$
Moreover, $|x_{,u}\times x_{,v}|=\sqrt{|x_{,u}|^2|x_{,v}|^2-\langle x_{,u},x_{,v}\rangle^2}$. Since $x$ is conformal, $|x_{,u}\times x_{,v}|=|x_{,u}||x_{,v}|$.
If we could see that $|x_{,u}\times x_{,v}|=\frac{|f|^2}{2}$, then
$$N(u,v)=\frac{x_{,u}\times x_{,v}}{|x_{,u}\times x_{,v}|}=$$
$$\frac{f\times \bar{f}}{|f|^2}=2\Im \frac{(f_2\bar{f_3},f_3\bar{f_1} ,f_2\bar{f_2} )}{|f|^2}$$
Let $G=St\circ N$. We have $$G=St(\frac{f\times\bar{f}}{|f|^2})=St(\frac{2\Im (f_2\bar{f_3},f_3\bar{f_1},f^1\bar{f_2})}{|f|^2})=$$
$$\frac{2\Im(f_2\bar{f_3})+i2\Im(f_3\bar{f_1})}{|f|^2-2\Im(f_1\bar{f_2})}$$
Please, help with this problem.
Edit: Thank you, James Cook. The problem is to verify the identity $g=G$.
Edit: A reference would be nice.
 A: First, we show that $|x_{,u} \times x_{,v}| = \frac{|f|^2}{2}$.  Notice that since $z = u+iv$, we have
$$x_{,u} = \mathfrak{R} \left(\frac{d x}{d z}\frac{\partial z}{\partial u}\right) = \mathfrak{R} f.$$
(In a similar manner, $x_{,v} = - \mathfrak{I} f$.)  So, we have
$|x_{,u} \times x_{,v}| = |x_{,u}| |x_{,v}| = |\mathfrak{R} f|^2$ by conformality.  Now we compute this:
\begin{align*} |\mathfrak{R} f|^2 & = \frac{1}{4} \left(f_1 + \overline{f_1}\right)^2 + \frac{1}{4} \left(f_2 + \overline{f_2}\right)^2 + \frac{1}{4} \left(f_3 + \overline{f_3}\right)^2 \\
& = \frac{1}{4} \left(f_1^2 + f_2^2 + f_3^2\right) + \frac{1}{2} \left(f_1 \overline{f_1} + f_2 \overline{f_2} + f_3 \overline{f_3} \right) + \frac{1}{4} \left(\overline{f_1^2} + \overline{f_2^2} + \overline{f_3^2}\right)\\
& = \frac{1}{2} |f|^2.
\end{align*}
To show that $G = g$, we have
\begin{align*} G & = \frac{2\mathfrak{I}(f_2 \overline{f_3}) + i 2\mathfrak{I}(f_3\overline{f_1})}{|f|^2 - 2 \mathfrak{I}(f_1\overline{f_2})}\\
& = \frac{2\left(\frac{f_2\overline{f_3} - \overline{f_2}f_3}{2i}\right) +2i \left(\frac{f_3\overline{f_1} - \overline{f_3}f_1}{2i}\right)}{f_1\overline{f_1} + f_2\overline{f_2} + f_3\overline{f_3} -2\left(\frac{f_1\overline{f_2} - \overline{f_1}f_2}{2i}\right)} \\
& = \frac{f_3(\overline{f_1}+i\overline{f_2}) - \overline{f_3} (f_1 + i f_2)}{f_1\overline{f_1} + f_2\overline{f_2} + f_3\overline{f_3} +if_1\overline{f_2} - i\overline{f_1}f_2}\\
&:= \frac{A}{B}.
\end{align*}
We would like to show that $\frac{A}{B} = \frac{f_3}{f_1-if_2}$, so it suffices to show that $A(f_1-if_2) = Bf_3$.  Now we compute
\begin{align*} A(f_1-if_2) & = f_3(f_1\overline{f_1} - i\overline{f_1}f_2 +i f_1\overline{f_2} + f_2\overline{f_2}) - \overline{f_3}(f_1^2 + f_2^2) \\
& = f_3(f_1\overline{f_1} - i\overline{f_1}f_2 +i f_1\overline{f_2} + f_2\overline{f_2}) - \overline{f_3}(-f_3^2) \\
& = f_3 \left(f_1\overline{f_1} - i\overline{f_1}f_2 +i f_1\overline{f_2} + f_2\overline{f_2} + f_3\overline{f_3}\right)\\
& = f_3 B
\end{align*}
where in the second line we used $f_1^2+f_2^2+f_3^2=0$.
