I found this problem in the book "Notes on set theory" by Yiannis Moschovakis; it's the x5.21 from the fifth chapter. You have to prove the following theorem:
for any three functions g: $Y$$\rightarrow$ $E$, h: $E$ $\times$$\mathbb{N}$$\times$$Y$ $\rightarrow$ $E$ and p: $\mathbb{N}$$\times$$Y$ $\rightarrow$ $Y$, there is exactly one function f: $\mathbb{N}$$\times$$Y$ $\rightarrow$ $E$ which satisfies the identities f (0, y) = g(y) and f(Sn, y) = h(f(n, p(n, y)), n, y) [where function S is the natural successor function].
Moschovakis gives a hint for the proof: define recursively a function F: $\mathbb{N}$ $\rightarrow$ $\mathbb{N}$$\times$($\mathbb{N}$ $\rightarrow$ $Y$) [where ($\mathbb{N}$ $\rightarrow$ $Y$) represents the function space i.e. the set of all functions from $\mathbb{N}$ to $Y$]; then the candidate for proving the theorem can be defined as f(n, y) = Second(F(n))(y) [the term 'Second' stands for the projection function of arity 2 whose value is the second component of its argument]. But I'm pretty sure that there is a typographical error in it because if F is defined in that way, the argument of the function obtained by composing the projection and F cannot be an element of $Y$ and it doesn't produce a value in $E$. So, any idea of the solution?
I suspect the following is what Moschovakis had in mind. By recursion we first define a sequence of functions from $Y$ to $E$:
$f_0$ is just $g$
$f_{n+1}$ is the function which takes value $h(f_n(p(n, y)),n, y)$ for each $y\in Y$
Our required function is then defined by $f(n, y) = f_n(y)$, and uniqueness can be proved by induction in the usual way.