Where $m_x$ and $\sigma_x$ are respectively the mean and standard deviation of $x(t)$, given an exact expression for $f_y(y)$:
$$ f_y(y)=\frac{1}{3(2\pi)^{1/2}\sigma_xy^{2/3}}\text{exp}(-\frac{(y^{1/3}-m_x)^2}{2\sigma^2_x}) $$
By integration, the following expressions for $m_y$ and $\sigma_y$ can be obtained:
$$ m_y=\sigma^3_x(3\lambda+\lambda^3) $$
and
$$ \sigma_y=\sigma^3_x(15+36\lambda^2+9\lambda^4)^{1/2} $$
where
$$ \lambda=\frac{m_x}{\sigma_x} $$
I do no know how to get the expectation and the variance of the above pdf, the answers are provided, can somebody show me how to do this integration?
You can make a couple substitutions reducing the integrals to a standard normal. The first one that comes to mind is $x = (y^{1/3}-m_x)/\sigma_x$. Then you get $$ \mathbb{E}[X] = \int_{-\infty}^\infty xe^{-x^2/2} dx $$ which can be solved using $u = -x^2/2$, and variance needs the second moment, $$ \int_{-\infty}^\infty x^2 e^{-x^2/2} dx $$ which requires going by parts differentiating $x$ and integrating $xe^{-x^2/2}$ as in the expected value...