Which of those functions are PRFs? Assume that $F: \{0,1 \}^n \rightarrow \{0,1 \}^n$ is PRF. Examine if the following functions are PRFs:
$$
\begin{align}
1. \, F_1(k,x) &amp;= F(k,x) \oplus x \\
2. \, F_2(k,x) &amp;= F\left(F(k,0^n), x\right) \\
3. \, F_3(k,x) &amp;= F\left(F(k,0^n), x\right)|| F(k,x)
\end{align}
$$
where $||$ denotes concatenation.

Attempt:
For $1.$ a standard distinguisher would be: $x_1 \oplus x_2 == y_1 \oplus y_2$, where $x_1, x_2$ are two inputs and $y_1, y_2$ are their outputs respectively. So $F_1$ is not a PRF.
For $2.$ I believe that $F(k,0^n) = k'$ acts as a key, so $F_2(k',x) = F(k',x)$ should still be a PRF.
For $3.$ I would say that the concatenation of two PRFs would still be a PRF, since you combine two pseudorandom outputs together to yield $F_3$'s output.
Can you verify my results?
Since @fgrieu was interested in question 2, I'm going to break with site policy and give a full answer for that part, even though this is almost certainly homework.
Theorem.
Let $F : \{0,1\}^n \times \{0,1\}^n \to \{0,1\}^n$ be a PRF. Then, $F_2 : \{0,1\}^n \times \{0,1\}^n \to \{0,1\}^n$, defined as
$$F_2(k,x) := F(F(k,0^n),x)$$
is also a PRF.
Before we give a formal proof, let's build some intuition why this should be the case. Note that the inner invocation of $F$ is not on some part of the input, but instead on a constant (i.e., $0^n$). This means that once we fix the key $k$, the key used by the outer invocation of $F$ for all evaluations of $F_2$ is a fixed key $k' := F(k,0^n)$. Since $k$ is not used elsewhere in the construction, $k'$ should, by the security of the underlying PRF, be indistinguishable from a uniformly random key $k''$, as long as $k$ is chosen uniformly at random. This would mean than an oracle evaluating $F_2(k,\cdot)$ and an oracle evaluating $F(k,\cdot)$ should in fact be indistinguishable. Since $F(k,\cdot)$ is already known to be indistinguishable from a uniformly chosen random function $f(\cdot)$ and indistinguishablity (in the asymptotic sense) is transitive, it would follow that $F_2$ must also be a PRF.
So let's formalize this intuition.
Proof. Let $A$ be an arbitrary PPT algorithm with
$$\Bigl|\Pr_k[A^{F_2(k,\cdot)}(1^n)=1] - \Pr_f[A^{f(\cdot)}(1^n)=1]\Bigr|=\epsilon(n).$$
We are looking to give a negligible upper bound for $\epsilon$. To do this we will prove a series of claims.
Claim 1. $\Bigl|\Pr\limits_k[A^{F(F(k,0^n),\cdot)}(1^n)=1] - \Pr\limits_{f}[A^{F(f(0^n),\cdot)}(1^n)=1]\Bigr| \leq \mathsf{negl}(n)$
Consider the following adversary $B$ against the PRF security of $F$.
Upon input $1^n$ and given access to an oracle, $B$ queries $0^n$ to the oracle an receives back a value $k'$. It then invokes $A$ on input $1^n$. Whenever $A$ sends a query $x$ to its oracle, $B$ responds by computing $y:=F(k',x)$. Eventually, $A$ will output a bit $b$ which $B$ also outputs.
It is easy to see that $$\Pr_k[B^{F(k,\cdot)}(1^n)=1] = \Pr_k[A^{F(F(k,0^n),\cdot)}(1^n)=1]$$ and $$\Pr_f[B^{f(\cdot)}(1^n)=1] = \Pr_f[A^{F(f(0^n),\cdot)}(1^n)=1].$$
Further, since $F$ is a secure PRF, it must hold that
$$\Bigl|\Pr_k[B^{F(k,\cdot)}(1^n)=1]-\Pr_f[B^{f(\cdot)}(1^n)=1]\Bigr|\leq \mathsf{negl}(n)$$
and the claim follows.
Claim 2. $\Pr\limits_{f}[A^{F(f(0^n),\cdot)}(1^n)=1] = \Pr\limits_{k}[A^{F(k,\cdot)}(1^n)=1]$
To see that this is the case, it is easiest to think of $f$ as being lazily sampled when queried. Since $f$ is only ever invoked on $0^n$, sampling $f$ is equivalent to simply sampling $f(0^n)$ once as a uniformly random value $k\in \{0,1\}$, which is identical to the right hand side.
Claim 3. $\Bigl|\Pr_{k}[A^{F(k,\cdot)}(1^n)=1] - \Pr_{f}[A^{f(\cdot)}(1^n)=1]\Bigr| \leq \mathsf{negl}(n)$
This claim is in fact just a restatement of the assumption that $F$ is a PRF and thus follows trivially.
Using the triangle inequality, we can conclude
\begin{align}
  \epsilon(n) =&amp;\quad \Bigl|\Pr_k[A^{F_2(k,\cdot)}(1^n)=1] - \Pr_f[A^{f(\cdot)}(1^n)=1]\Bigr|\\
  \leq&amp;\quad\Bigl|\Pr\limits_k[A^{F(F(k,0^n),\cdot)}(1^n)=1] - \Pr\limits_{f}[A^{F(f(0^n),\cdot)}(1^n)=1]\Bigr|\\ 
&amp;+ \Bigl|\Pr\limits_{f}[A^{F(f(0^n),\cdot)}(1^n)=1] - \Pr\limits_{k}[A^{F(k,\cdot)}(1^n)=1]\Bigr|\\ 
&amp;+ \Bigl|\Pr_{k}[A^{F(k,\cdot)}(1^n)=1] - \Pr_{f}[A^{f(\cdot)}(1^n)=1]\Bigr|\\
&amp;\leq 2\cdot\mathsf{negl}(n)
\end{align}
and the theorem statement immediately follows.$\tag*{$\square$}$