If one root of the quadratic equation

Since, $x=\frac{-1}{3}$ is a root of the quadratic equation $2 x^{2}+2 x+k=0$, then, it must satisfies the equation.

$2\left(-\frac{1}{3}\right)^{2}+2\left(-\frac{1}{3}\right)+k=0$

$\Rightarrow 2\left(\frac{1}{9}\right)-\frac{2}{3}+k=0$

$\Rightarrow \frac{2}{9}-\frac{2}{3}+k=0$

$\Rightarrow \frac{2-6+9 k}{9}=0$

$\Rightarrow-4+9 k=0$

$\Rightarrow 9 k=4$

$\Rightarrow k=\frac{4}{9}$

Hence, the value of $k$ is $\frac{4}{9}$.