Solution
= Solution
Apply the <S-m-n theorem> to $g$. It gives a total computable function $h$ such that
$$
f_{h(x),1}(y)\simeq g(x,y).
$$
The <recursion theorem> supplies an index $e$ satisfying $f_{e,1}\simeq f_{h(e),1}$. Consequently
$$
\boxed{f_{e,1}(y)\simeq g(e,y)\quad\text{for every }y\in\mathbb N_0.}
$$