Proving correctness of pseudorandom generator construction given existing pseudorandom generator
Call your original ($k$-to-$3k$ bit) PRG $G$ and your construction $G'$.

Let $\mathcal{U}_t$ denote the uniform distribution on $\{0,1\}^t$. Then $G'$ is a PRG as long as the distributions $\mathcal{U}_{3k}$ and $G'(\mathcal{U}_{2k})$ are indistinguishable with effort polynomial in $k$; see this reference.

Distribution $G'(\mathcal{U}_{2k})$ is just $G(\mathcal{U}_k) \oplus G(\mathcal{U}_k)$, where the two occurrences of $\mathcal{U}_k$ in the latter expression are independent. By the PRG property of $G$ applied to the first term, the distribution is indistinguishable from $\mathcal{U}_{3k} \oplus G(\mathcal{U}_k)$.

Since $\mathcal{U}_t \oplus \mathcal{D}$ is distributed identically to $\mathcal{U}_t$, for any independent distribution $\mathcal{D}$, we get the desired result.



I find this question a bit odd though. If all you want is a $2k$-to-$3k$ PRG constructed from a $k$-to-$3k$ PRG, then what about the following simpler construction: given $2k$ bits, throw away the first half and run the second half through the $k$-to-$3k$ PRG?