Well, $f''f^2 +xf' ^3+2ff'^2 =0$ modulo 3 for $f=\prod(1-x^m)=\sum_{n\in \mathbb{Z} } (-1)^nx^{n(3n+1)/2}$ may be quickly seen as follows.

Differentiating the power series for $f$ and expanding the brackets we see that we should prove that
$$
\sum_{a(3a+1)/2+b(3b+1)/2+c(3c+1)/2=n} 
(-1)^{a+b+c} \left(a(a-2)-abc+2ab\right) 
$$
is divisible by 3. Multiplying the sum by 3 and using cycling shift of variables we reduce it to proving that the sum
$$
\sum_{a(3a+1)/2+b(3b+1)/2+c(3c+1)/2=n} 
(-1)^{a+b+c} \left( a(a-2)+b(b-2)+c(c-2)-3abc+2ab+2bc+2ac\right) 
$$
is divisible by 9.

The summand is $(a+b+c) (a+b+c-2)-3abc$. If $a=b=c$ this is $9a^2 - 3(a^3+2a)$, divisible by 9. Other triples partition by permuting the variables onto 3-tuples and 6-tuples, so the sum of $3abc$ is of course divisible by 9. As for $a+b+c$, it is congruent to $2n$ modulo 3, thus unless $n=3t+2$ the expression $(a+b+c) (a+b+c-2)$ is divisible by 3, as we need. 

It remains to show that for $n=3m+2$ the sum of $(-1)^{a+b+c}$ over our triples is divisible by 9. In other words, the coefficient of $x^{3m+2} $ in $f^3$ must be divisible by 9. We have 
$$
f^3=\prod (1-x^k)^3 =\prod (1-x^{3k}+3(x^k-x^{2k})).
$$
Expanding the brackets and reducing modulo 9 we should prove that the expression
$$
[x^{3m+2} ] 3 f(x^3) \sum_k \frac{x^k-x^{2k} } {1 - x^{3k} }
$$
is divisible by 9.
But in the latter sum $\sum_k (x^k-x^{2k}+x^{4k}-x^{5k}+\dots) $ all coefficients of powers $x^{3s+2} $ do cancel, since if $3s+2=kr$, the guys $x^{kr} $ and $x^{rk} $ go with different signs.