$x=\frac{2}{3}$ is one of the zero of $3 x^{3}+16 x^{2}+15 x-18$ Now, we have

$\mathrm{x}=\frac{2}{3}$

$\Rightarrow \mathrm{x}-\frac{2}{3}=0$

Now, we divide $3 x^{3}+16 x^{2}+15 x-18$ by $x-\frac{2}{3}$ to find the quotient

So, the quotient is $3 x^{2}+18 x+27$

Now,

$3 x^{2}+18 x+27=0$

$\Rightarrow 3 x^{2}+9 x+9 x+27=0$

$\Rightarrow 3 x(x+3)+9(x+3)=0$

$$
\begin{aligned}
&\Rightarrow(x+3)(3 x+9)=0 \\
&\Rightarrow(x+3)=0 \text { or }(3 x+9)=0 \\
&\Rightarrow x=-3 \text { or } x=-3
\end{aligned}
$$