Let $\ S\ $ be the set defined by the equations
$\ {x}^{2}+{y}^{2}+{z}^{4}=3\ $ and $\ {x}^{3} - {y}^{3} + z(1+xy) =2 \ $
Let $\ f(x,y,z)=e^{x+yz} + y{x}^{3}.\ $
Show that, for $\ P=(1,1,1)\ $and some $\epsilon > 0,\ $
$$\ M=S \cap {B}_{P(\epsilon)}\ $$
is a manifold, where $\ {B}_{P(\epsilon)}\ $ is the open ball of radius $\epsilon $ around $\ P.$
Is the set $\ M\ $ consist of the following three equations where I need to use the implicit function theorem to show to be a manifold:
$$\ {x}^{2}+{y}^{2}+{z}^{4}=3,\ $$ $$\ {x}^{3} - {y}^{3} + z(1+xy) =2\ $$ and $$\ {(x-1)}^{2}+{(y-1)}^{2}+{(z-1)}^{2}= \epsilon^{2},\ $$
I don't see the purpose of your function $f$. If $S$ is the set defined by the equations $$x^2+y^2+z^4 = 3 \quad\mbox{and}\quad x^3-y^3 + z(1+xy) = 3,$$then $(1,1,1) \not\in S$ because $1^3 - 1^3 + 1(1+1\cdot 1) = 2 \neq 3$. So there is $\epsilon$ small enough such that $S \cap B(P,\epsilon) = \varnothing$. So the question has no answer as it is written.