Let $F(k,x)$ be a secure PRF over $(\mathcal{K},\mathcal{X},\mathcal{Y})$ where $\mathcal{K} = \mathcal{X} = \mathcal{Y} = \{0,1\}^n$.
Let $F'(k, x) = F(F(k, 0^n), x) \; \Vert \; F(k, x)$.
$a \; \Vert \; b$ means $a$ concatenated to $b$.
How can I show that $F'$ is not a secure PRF? I've been trying to build an adversary that could find some pattern but I couldn't. My best guess was to initially use $x = 0^n$, so the output would be something like $F(k', 0^n) \; \Vert \; k'$ (where $k' = F(k, 0^n)$, which is the key of the left side) and I would be able to use $k'$ for the next calculations, but it didn't seem to be enought to find any pattern in the sequence of outputs.