Assuming that each normal distribution is independent, we can use an identity to derive the covariance matrix of $\mathbf{x}$. Observe the identity below.
Let $Y$ and $Z$ be independent normally distributed random variables. If $Y\sim N(\mu_Y, \Sigma_Y)$ and $ Z\sim N(\mu_Z, \Sigma_Z)$ then
$$X=AZ+BY\sim N(A \mu_Y + B \mu_Z , A \Sigma_Y A^T + B \Sigma_Z B^T)$$
where $A$ and $B$ are simple bivariate linear transformations.
Therefore, using the identity we can simply solve by plugging and chugging. Just note that the positive definite form of the covariances can be simplified when the transformation ($A$ or $B$) is only a scalar then the covariance is only the scalar squared multiplied by the covariance matrix, e.g. $\alpha \Sigma_Y \alpha^T = \alpha^2 \Sigma_Y$
$$X\sim N(\mu_X, \Sigma_X) $$
$$\mu_X= \frac{1}{3}\times 0 +\frac{1}{3} \times\left(\begin{array}{c}-6 \\-6\end{array}\right)+\frac{1}{3} \times \left(\begin{array}{c}-4 \\-4\end{array}\right)$$
$$\Sigma_X= \frac{1}{3^2} \left(\begin{array}{cc}1 & 0 \\0 & 1\end{array}\right) + \frac{1}{3^2} \left(\begin{array}{cc}1 & 0.9 \\0.9 & 1\end{array}\right)+ \frac{1}{3^2} \left(\begin{array}{cc}1 & -0.9 \\-0.9 & 1\end{array}\right)$$
By adding all the terms for $\Sigma_X$, you can identify correlation by just observing the off-diagonals. From the identity, we also identified an expression for the mean.
