$\newcommand{\cov}{\operatorname{cov}}$Let $X \sim \operatorname{Uniform}[-1,1]$ and assume that $Y = -\frac{2}{3} + X^2+ V, V\mid X \sim N(0, \sigma^2)$. Calculate $\cov(X, Y)$.
What I did was substitute in the expression for $Y$ and then use properties of covariance to expand it:
\begin{align} \cov(X, Y) & = \cov(X, -\frac{2}{3} + X^2+ V) = \cov(X, -2/3) + \cov(X, X^2) + \cov(X, V) \\[10pt] & = \cov(X, X^2) + \cov(X, V) \end{align}
But how do I work out $\cov(X, X^2)$ and $\cov(X, V)$?
In general $\operatorname{Cov}(Z_1, Z_2) = \mathbb{E}[Z_1 Z_2] - \mathbb{E}[Z_1] \mathbb{E}[Z_2]$. $$\operatorname{Cov}(X,X^2) = \mathbb{E}[X^3] - \mathbb{E}[X] \mathbb{E}[X^2] = \cdots$$ then use the fact that $X \sim \operatorname{Unif}[-1,1]$ to compute the expectations.
Next, $$\operatorname{Cov}(X,V) = \mathbb{E}[XV] - \mathbb{E}[X] \mathbb{E}[V].$$ If the notation in your post is intended to convey that $X$ and $V$ are independent, then this covariance can be shown to be zero. (Why?)