You have $y'(x)+3y(x)=8\Rightarrow \mathbb{e}^{3x}y'(x)+3\mathbb{e}^{3x}y(x)=8\mathbb{e}^{3x}\Rightarrow \mathbb{e}^{3x}y'(x)+\left(\mathbb{e}^{3x}\right)'y(x)=8\mathbb{e}^{3x}\Rightarrow$
$\left(\mathbb{e}^{3x}y(x)\right)'=8\mathbb{e}^{3x}\Rightarrow {\Large\int}\left(\mathbb{e}^{3x}y(x)\right)'\;\mathbb{d}x={\Large\int}8\mathbb{e}^{3x}\;\mathbb{d}x\Rightarrow \mathbb{e}^{3x}y(x)=\frac{8}{3}\left(\mathbb{e}^{3x}\right)+c\Rightarrow $ *for $c\in\mathbb{R}$
$y(x)=\frac{8}{3}+\frac{c}{\mathbb{e}^{3x}}$, now since $y(0)=0\Rightarrow c=-\frac{8}{3}$ and $y(x)=\frac{8}{3}-\frac{8}{3\mathbb{e}^{3x}}$.