Number of integer partitions modulo 3
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.
The function $\,h(x) := \prod_{n=1}^\infty (1 - x^n)\,$ is known as a Ramanujan theta function. It is essentially the Dedekind $\eta$ function. The connection is $\,f(q) := q h(q^{24}) = \eta(24\tau)\,$ where $\,q=\exp(2\pi i \tau).\,$ Differentiating we get $\, dq = (2\pi i)\, q\, d\tau,\,$ and $$ \frac{d}{d\tau} f(q) = (2\pi i)\, q\, f'(q).$$ Since all powers of $\,q\,$ in $\,f(q)\,$ are of the form $\,q^{24n+1}\,$ then the congruence $\,f'(q) - f(q) \equiv 0 \pmod 3\,$ holds.
If we use the original $\,h(x)\,$ then there is similar but more complicated result using modulo $3$ exponents. Let$\, A = A(x) := h(x)\,$ and let $\,A_0\,$ be all the terms of $\,A\,$ where the exponent of $\,x\,$ is $0$ modulo $3$ and similarly for $\,A_1\,$ and $\,A_2\,$ so that $\,A = A_0+A_1+A_2.\,$ This is the trisection of the power series $\,A.\,$ Refer to my essay A Multisection of q-Series for the nice identity $$ 0 = A_2 A_0^2 + A_0 A_1^2 + A_1 A_2^2. \tag{1}$$ Now let $\, B = B(x) := A'(x)\,$ and let $\,B_0,B_1,B_2\,$ be the trisection of $\,B.\,$ It is easy to show that $$ q\,B_0 \equiv A_1,\quad q\,B_1 \equiv -A_2, \quad B_2 \equiv 0 \pmod 3.$$ Now let $\, C = C(q) := A''(x)\,$ and let $\,C_0,C_1,C_2\,$ be the trisection of $\,C.\,$ It is easy to show that $$ q^2\,C_0 \equiv -A_2, \quad C_1 \equiv C_2 \equiv 0 \pmod 3.$$ When we make these substitutions in the expression $$ h''(x) h(x)^2 + x\, h'(x)^3 + 2\,h(x)h'(x)^2 \tag{2}$$ and reduce modulo $3$ we get the nice identity $(1)$.