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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 $$\ {(x-1)}^{2}+{(y-1)}^{2}+{(z-1)}^{2}= \epsilon^{2},\ $$

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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.

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  • $\begingroup$ I rechecked the typing of my question. It seems that the second equation should equal to 2 instead of 3. I edited the question to the correct change $\endgroup$ – Seth Mai Apr 6 '18 at 15:13

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