If $\Lambda$ is a lattice in $\Bbb C$ the map $$z\mapsto (\wp(z),\wp'(z))$$ is a parametrisation of the complex points of the elliptic curve $$E:\qquad y^2=4x^3-g_2x-g_3$$ where $g_2$ and $g_3$ depend on $\Lambda$. It induces a group isomorphism $Φ$：$\Bbb C/\Lambda\to E(\Bbb C)$ : $t\mapsto (\wp(t),\wp'(t))$ .
To show this is isomorphism as Rieman surface, it is enough to prove $Φ$ is local analytic isomorphism ( because $Φ$ is bijective).
So, we only need to check
Φ*(dx/y)＝$d\wp(z)/\wp'(z)$＝$z$・・・①.
Once ① is proved, this is holomorphic and non vanishing at every point of domain, so $Φ$ is local analytic function.
My question: Why first and second equal of ① holds?
Formally, we have $\Phi=(\wp,\wp')$, and people use the letters $x$, $y$ like $\Phi(z)=(\wp(z),\wp'(z))\ "="\ (x,y)$.
Note that $x$ stays also abusively for a "function name", for the projection onto the first component of a point $(x,y)\in E(\Bbb C)$. (To be pedant, we need to introduce $\pi_1$ which takes $(x,y)$ to $x$. Then trying to use $\pi_1$ for, say" a "polynomial expression in $x,y$... to complicated, we just use the polynomial expression, "knowing what is meant". At some point we no longer care, and write $x$ for $\pi_1$.)
So we compute the pull back of $x$, $dx$, $y$, $dx/y$ (formally): $$ \begin{aligned} \Phi^*(x) &= x\circ \Phi\ ,\\ \Phi^*(y) &= y\circ \Phi\ ,\\[2mm] \Phi^*(x)(z) &= x\circ \Phi(z)=x\circ(\wp(z),\wp'(z))=\wp(z)\ ,\\ \Phi^*(y)(z) &= y\circ \Phi(z)=y\circ(\wp(z),\wp'(z))=\wp'(z)\ , \text{ so formally}\\[2mm] \Phi^*(x) &= \wp(z)\ ,\\ \Phi^*(y) &= \wp'(z)\ ,\\[2mm] \Phi^*(dx) &= d\Phi^*(x)=d\wp(z)=\wp'(z)\; dz\ ,\\ \Phi^*\left(\frac{dx}y\right) &= \frac{\wp'(z)\; dz}{\wp'(z)}=dz\ . \end{aligned} $$ (The invariant differential $\displaystyle\frac 1y\; dx$ is thus mapped contravariantly into the invariant differential $dz$. The first invariance is w.r.t. the addition on the elliptic curve $E(\Bbb C)$, the second w.r.t. the addition on $\Bbb C$.)