TITLE: de Rham cohomology of $x^3+y^3 +z^3+c \,xyz= 0$ find representatives
QUESTION [6 upvotes]: The equation $x^3+y^3 +z^3+c\, xyz= 0$ defines a non-singular elliptic curve $X$ in $\mathbb{C}P^2$ projective space.
In fact, how do we prove it has genus 1 both as an algebraic curve and as a Riemann surface?
Since $H^1(X,\mathbb{C})\simeq \mathbb{C}^2$ what are the explicit representatives of the de Rham cohomology? There should be two of them.

This question is (verbatim) on Math.Stackexchange unanswered even with a bounty.  

despite a change of variables, I am guessing it might be best to maintain the symmetry between $x,y,z$ if possible.
I am asking for explicit sections in a manner that might have been done in the 19th century (but could be done in many situations today).  they should correspond to classical functions in some way.
there's an underlying question of whether the "cohomology" of this curve as a topological space matches up with the sheaf cohomology or de Rham cohomology. Especially when the curve becomes singular.
I'm also looking for examples where they disagree

REPLY [4 votes]: A possible method for getting explicit representatives involves Griffiths's Residues. More in general, let $X \subset \mathbb{P}^{n+1}$ a smooth hypersurface of degree $d$.
Denote by $H^n_{\textrm{prim}}(X)$ the primitive subspace of the n-th cohomology of $X$, that is  $$ H^n_{\textrm{prim}}(X)= \textrm{Ker} ( \lambda: H^n(X) \to H^{n+2}(X) ),$$ $$\lambda(\alpha)= \alpha \cup c_1 (\mathcal{O}_X(1)).$$
The (restriction on) the Hodge filtration $F^p H^n_{\textrm{prim}}(X)$ can be identified with $$H^0(\Omega^{n+1}_{\mathbb{P}}(n-p+1)(X)) / d H^0((\Omega^{n}_{\mathbb{P}}(n-p)(X)).$$
Now, if $$\omega= \sum_i (-1)^i x_i dx_0 \wedge \ldots d\hat{x_i} \wedge \ldots dx_{n+1}$$ write $$H^0(\Omega^{n+1}_{\mathbb{P}}(k))=\big\{ \frac{g \omega}{f^k}\ | \ g \in \mathbb{C}[x_0, \ldots, x_{n+1}]_{kd-n-2} \big \} $$
and use this identification in the quotient above.
One has that if $R_f:= \mathbb{C}[x_0, \ldots, x_{n+1}]/J_f$ with $J_f$ being the ideal generated by the partial derivatives of $f$ then $$H^{n-p,p}_{\textrm{prim}}(X):= H^n_{\textrm{prim}}(X) \cap H^{n-p,p}(X) \cong (R_f)_{(p+1)d-n-2},$$
where the maps sends the class of $g\omega/f^k$ to the class of $g$.
Now this is all a very nice bit of theory, but in this case the situation is very simple. For a curve we have $H^1(C, \mathbb{C}) \cong H^{1,0}(C) \oplus H^{0,1}(C)$ and moreover $H^1_{\textrm{prim}}(C) \cong H^1(C, \mathbb{C})$ (and the same for the subspaces of course).
The degree of the equation is three, therefore one has $H^0(K_C) \cong (R_f)_0$ and $H^1(\mathcal{O}_C) \cong (R_f)_3$. The former is generated by 1 (it is canonically identified with $\mathbb{C}$). For the latter, your equation for $C$ implies that you can pick as generator for $(R_f)_3 \cong \mathbb{C}\cong \langle z^3 \rangle$. Note that $R_3$ is the socle of the Jacobian ring, that is $(R_f)_t=0$ for $t>3$. 
Moreover (in the non weighted case) if $\rho$ is the degree corresponding to the socle of this (artinian, Gorenstein) ring one has $R_a \cong (R_{\rho-a})^*$, and this is in turn just another expression of Serre Duality.
Plugging these polynomials in the quotient above, one gets an explicit expression as desired.