Given the following mock theta functions of order $3$,found in this wikipedia article
$f(q)=\sum_{n=0}^{\infty} \frac{q^{n^2}}{(-q;q)^2_{n}}$,$\phi(q)=\sum_{n=0}^{\infty} \frac{q^{n^2}}{(-q^2;q^2)_{n}}$,$\omega(q)=\sum_{n=0}^{\infty} \frac{q^{2n(n+1)}}{(q;q^2)^2_{n+1}}$ and $\nu(q)=\sum_{n=0}^{\infty} \frac{q^{n(n+1)}}{(-q;q^2)_{n+1}}$
where $q=e^{2\pi i\tau}$,$|q|\lt1$ and $(a;q)_n=\prod_{j=0}^{n-1}(1-aq^j)$
I propose the conjecture that the following relations hold
$\Big(\phi(q)-1\Big)^2+\Big(f(q^2)-1\Big)=2\sum_{k=1}^{\infty}\Big(\sum_{n=1}^{\infty} \frac{q^{(n-1)^2+2k(n-1)+2k^2}}{(-q^2;q^2)_{k} (-q^2;q^2)_{n+k-1}}\Big)\tag1$
and
$\Big(\nu(q)-\frac{1}{(1+q)}\Big)^2+\Big(\omega(-q)-\frac{1}{(1+q)^2}\Big)=2\sum_{k=1}^{\infty}\Big(\sum_{n=1}^{\infty} \frac{q^{n^2+(2k-1)n+2k^2}}{(-q;q^2)_{k+1} (-q;q^2)_{n+k}}\Big)\tag2$
How do we prove the conjecture?
Note:comparing coefficients on the RHS and LHS of $(1)$ and $(2)$ respectively,we observe that the coefficients are equal
and the identities are both special cases of the more general identity
$\Big(\sum_{n=1}^{\infty} \frac{x^n z^{n^2}}{(a;q)_{n}}\Big)^2+\sum_{n=1}^{\infty} \Big(\frac{x^n z^{n^2}}{(a;q)_{n}}\Big)^2=2\sum_{k=1}^{\infty}\Big(\sum_{n=1}^{\infty} \frac{x^{n+2k-1} z^{(n-1)^2+2k(n-1)+2k^2}}{(a;q)_{k} (a;q)_{n+k-1}}\Big)\tag3$
with complex number $x$ and $z$,$z\lt1$whereby a transformation($(a;q)_n\rightarrow (a;q)_{n+c}$ or $(a;q)_n\rightarrow (a;q)_{cn}$) on the LHS results in an equivalent transformation($(a;q)_k\rightarrow (a;q)_{k+c}$ or $(a;q)_k\rightarrow (a;q)_{ck}$) on the RHS for natural number $c$
The close resemblance of the summation for the pair $f(q^2)$, $\phi(q)$ and also for the pair $\nu(q)$, $\omega(-q)$ suggests that this is a good clue. In all four cases we next omit the $n=0$ term for the next step.
Using elementary algebra, let $S_1:=a_1+a_2+\dots,\; S_2:=a_1^2+a_2^2+\dots$, then $S_1^2-S_2=2\sum_{i\lt j}a_ia_j$ and $S_1^2+S_2=2\sum_{i\le j}a_ia_j$. The two equations are special cases of this general identity using the obvious choices for $a_1,a_2,\dots$. For example, the equation (1) uses $a_n = q^{n^2}/(-q;q)^2_n$.