Show $x^y$ is a primitive recursive function
As I wrote in the comments, you haven't actually proved anything.  If you want to prove that $x^y$ is p.r., you need to write it in the scheme form, and then you need to prove that indeed $x^y = f(y,x)$ for your defined $f$ for all $x$ and $y$ (by double induction).

Consider the following definition of $f(y,x) = x^y$:

$f(0,x) = s(z(x))$ (i.e., the successor of the constant zero function).

$f(y+1, x) = u^3_1(u^3_1(x^y, x, y) \cdot x, x, y)$.

I assume now that multiplication, $x \cdot y$ and wrote it infix, and obviously $x^y = f(y,x)$.  The function $u^a_b$ is the $a$-ary projection of $b$.

(Ps. if this doesn't work for you, please post your scheme you want to use, otherwise we cannot do else then guess.)