How to show that $F(x)=F(xq)+xq F(xq^2)$ Define$$\begin{align*}F(x) & =1+\dfrac {xq}{1-q}+\dfrac {x^2q^4}{(1-q)(1-q^2)}+\dfrac {x^3q^9}{(1-q)(1-q^2)(1-q^3)}+ \cdots\\ & =1+\sum\limits_{n=1}^{\infty}\dfrac{x^nq^{n^2}}{(q;q)_{n}}\end{align*}$$
Then $F(x)$ also satisfies $F(x)=F(xq)+xq\cdot F(xq^2)$.

Question: How would you prove $F(x)=F(xq)+xq\cdot F(xq^2)$?

I attempted substituting $x=xq$ into $F(x)$ to get$$\begin{align*}F(xq) & =1+\dfrac {xq^2}{1-q}+\dfrac {x^2q^6}{(1-q)(1-q^2)}+\dfrac {x^3q^{12}}{(1-q)(1-q^2)(1-q^3)}+\cdots\\ & =1+\sum_{n=1}^{\infty}\dfrac {x^nq^{n^2+n}}{(q;q)_n}\end{align*}$$
And$$\begin{align*}F(xq^2) & =1+\dfrac {xq^3}{1-q}+\dfrac {x^2q^8}{(1-q)(1-q^2)}+\dfrac {x^3q^{15}}{(1-q)(1-q^2)(1-q^3)}+\cdots\\ & =1+\sum\limits_{n=1}^{\infty}\dfrac {x^nq^{n^2+2n}}{(q;q)_n}\end{align*}$$
So $xq\cdot F(xq^2)=xq+xq\sum\limits_{n=1}^{\infty}\dfrac {x^nq^{n^2+2n}}{(q;q)_n}$. However, adding them together,$$F(xq)+xq\cdot F(xq^2)=1+\sum\limits_{n=1}^{\infty}\dfrac {x^nq^{n^2+n}}{(q;q)_n}+xq+xq\sum\limits_{n=1}^{\infty}\dfrac {x^nq^{n^2+2n}}{(q;q)_n}$$Which I don't think equals $F(x)$.

Using the convention $(q;q)_0 = 1$, we can write
$$F(x) = \sum_{n = 0}^\infty \frac{x^n q^{n^2}}{(q;q)_n}$$
So 
$$F(xq) = \sum_{n = 0}^\infty \frac{x^n q^{n^2 + n}}{(q;q)_n}$$
and
$$xq F(xq^2) = xq\sum_{n = 0}^\infty \frac{x^n q^{n^2+2n}}{(q;q)_n} = \sum_{n = 0}^\infty \frac{x^{n+1}q^{n^2+2n+1}}{(q;q)_n} = \sum_{n = 1}^\infty \frac{x^nq^{n^2}}{(q;q)_{n-1}} = \sum_{n = 1}^\infty \frac{x^nq^{n^2}}{(q;q)_n}(1-q^n)$$
Thus
$$F(xq) + xqF(xq^2) = 1 + \sum_{n = 1}^\infty \frac{x^nq^{n^2}}{(q;q)_n}(q^n + (1-q^n)) = 1 + \sum_{n = 1}^\infty \frac{x^nq^{n^2}}{(q;q)_n} = F(x)$$
