Let $G \approx \langle a,b \mid b^{-1}a^3b = a^5 \rangle$. Let $H$ be a finite group and $\varphi:G \to H$ be a homomorphism.
Then $g = a^{-1}b^{-1} a^{-1}bab^{-1}ab \in \ker \varphi$.
My attempt: I haven't gotten very far, but here are some observations:
Let $s = \varphi(a)$ and $t = \varphi(b)$. Then $s^n = 1$ and $t^k =1$ for some $n,k$. Without loss of generality, let $n,k$ be the smallest such numbers. $n$ cannot be a multiple of $3$ or $5$:
If $n=5m$ for some integer $m$, then $1=s^{5m}=t^{-1}s^{3m} t$, and thus $1 = s^{3m}$, contradicting the minimality of $n$. Similarly, $n$ is not a multiple of $3$.
Also, we have that $ag$ is conjugate to $a$ (somebody in chat noticed this).
I'm thankful for any insight.
You have defined $s=\varphi(a)$ and $n = o(s)$ (the order of $s$). So $o(s^3) = s/\gcd(n,3)$ and $o(s^5) = n/\gcd(n,5)$. But $s^3$ and $s^5$ are conjugate in $H$, so $o(s^3)=o(s^5)$, which can only happen if $\gcd(n,3) = \gcd(n,5) = 1$.
But then $s^3$ generates $\langle s \rangle$, so $s$ is a power, $(s^3)^k$ say, of $s^3$, and hence $$\varphi(b^{-1}ab) = \varphi(b^{-1})(s^3)^k\varphi(b) = \varphi(b^{-1}a^{3k}b) = \varphi(a^{5k})=\varphi(a)^{5k},$$ and so $\varphi(b^{-1}ab)$ commutes with $\varphi(a)$, and the result follows.