Since $y'=1$,
$$
\begin{align}
\left\langle\delta_a,(2\phi y+\phi'y^2)'\right\rangle
&=\left\langle\delta_a,2\phi+2y\phi'+\phi''y^2+2y\phi'\right\rangle\\
&=\left\langle\delta_a,2\phi\right\rangle
\end{align}
$$
since $\langle\delta_a,y\psi(y)\rangle=0$ for any $\psi$.
So, yes, your answer seems correct.
Take $\varphi_1 \in \mathcal{D}'(\mathbb{R}^{n-1})$ and $\varphi_2 \in \mathcal{D}'(\mathbb{R}).$
If $\varphi_2(0)=0$ then there exists $\psi \in \mathcal{D}'(\mathbb{R})$ such that $\varphi_2(x) = x \, \psi(x)$ giving
$$
\langle u, \varphi_1 \otimes \varphi_2 \rangle
= \langle u, \varphi_1 \otimes x \psi \rangle
= \langle x_n u, \varphi_1 \otimes \psi \rangle
= \langle 0, \varphi_1 \otimes \psi \rangle
= 0.
$$
Otherwise, take $\rho \in \mathcal{D}'(\mathbb{R})$ such that $\rho(0)=1,$ and set $\hat{\varphi}_2 = \varphi_2 - \varphi_2(0) \rho.$ Then $\hat{\varphi}_2(0) = 0$ so $\langle u, \varphi_1 \otimes \hat{\varphi}_2 \rangle = 0.$ Thus
$$
\langle u, \varphi_1 \otimes \varphi_2 \rangle
= \langle u, \varphi_1 \otimes (\hat{\varphi}_2 + \varphi_2(0) \rho) \rangle
= \langle u, \varphi_1 \otimes \hat{\varphi}_2 \rangle + \langle u, \varphi_1 \otimes \varphi_2(0) \, \rho \rangle \\
= \langle u, \varphi_1 \otimes \rho \rangle \varphi_2(0)
.
$$
Now, define $v_\rho \in \mathcal{D}'(\mathbb{R}^{n-1})$ by
$\langle v_\rho, \varphi_1 \rangle
= \langle\langle u(x',x_n), \rho(x_n)\rangle, \varphi_1(x')\rangle
.$
Then
$$
\langle u, \varphi_1 \otimes \rho \rangle \varphi_2(0)
= \langle\langle u(x',x_n), \rho(x_n)\rangle, \varphi_1(x')\rangle \, \langle \delta, \varphi_2 \rangle
= \langle v_\rho, \varphi_1\rangle \langle \delta, \varphi_2 \rangle
= \langle v_\rho \otimes \delta, \varphi_1 \otimes \varphi_2 \rangle.
$$
Thus,
$
\langle u, \varphi_1 \otimes \varphi_2 \rangle
= \langle v_\rho \otimes \delta, \varphi_1 \otimes \varphi_2 \rangle
$
and by uniqueness, we must have $u = v_\rho \otimes \delta.$
Best Answer
You don't need to integrate here, just use the definition of $\delta$:
$$\langle fH', \varphi \rangle = \langle H', f\varphi \rangle = \langle \delta, f\varphi \rangle = (f\varphi)(0) = f(0) \varphi(0).$$