The Wolfram Functions Site gives this partial differentiation equation w.r.t. $g_2,g_3$ for the Weierstrass sigma function
$$ z\frac{\partial \sigma(z; g_2, g_3)}{\partial z} - 4g_2\frac{\partial \sigma(z; g_2, g_3)}{\partial g_2} - 6g_3\frac{\partial \sigma(z; g_2, g_3)}{\partial g_3} - \sigma(z; g_2, g_3) = 0. $$
I don't get how to use the homogeneity property of sigma to prove this. Euler's theorem on homogeneous functions does not directly apply since sigma is not homogeneous in $(z,g_2,g_3)$.
If I assume that the power series expansion of sigma is a sum of terms of the form (where $c$ is a constant and $a$ and $b$ are integers) $\, c g_2^a g_3^b z^{1+4a+6b},\,$ then a simple computation shows the PDE is true for the term. This reduces the problem to proving this property of sigma.