# Prove that $xy^2+y^3z^4+z^5x^6=1$ has a solution in an open neighborhood about the point $x_0=(0,1,-1)$

Prove with implicit function theorem that $$xy^2+y^3z^4+z^5x^6=1$$ has a $$C^1$$ solution with a form of $$(x,g(x,z),z)$$ in an open neighborhood about the point $$x_0=(0,1,-1)$$.

What I have gotten so far:

Let $$f(x,y,z)=xy^2+y^3z^4+z^5x^6-1$$.

Now $$f(0,1,-1)=0\cdot1^2+1^3\cdot(-1)^4+(-1)^5\cdot0^6-1=0$$.

Also $$D_yf(x,y,z)=2xy+3z^4y^2$$ and $$D_yf(0,1,-1)=2\cdot0\cdot1+3\cdot(-1)^4\cdot1^2=3\ne0$$.

Now I can conclude with implicit function theorem that the function has a $$C^1$$ solution in an open neighborhood about the point $$x_0=(0,1,-1)$$ but how do I know it takes the form $$(x,g(x,z),z)$$?

• Hint: consider when the condition does not hold. Dec 29, 2018 at 9:19
• IFT specifically correlates the non-vanishing of the partial derivatie with respect to $y$ with the ability to "implicitly" solve $y$ locally as a $C^1$-function of the other coordinates. If you want to "solve" $z$ in terms of $x$ and $y$ instead of $y$, then you need to check that $D_zf$ does not vanish. Dec 29, 2018 at 9:30
• @JyrkiLahtonen Hi! Firstly, thank you for answering. However, I'm afraid I don't quite understand. How do I know I have to solve z in terms of x and y? Would the same statement hold if I choose to solve x in terms of y and z?
– Joe
Dec 29, 2018 at 10:24
• If you want to solve $x$, then you need to check that $D_x f$ does not vanish at the point of interest. Dec 29, 2018 at 10:27
• In a sense IFT is about linearizing the solution set (locally and approximately). So, when teaching, I emphasize the analogy with linear equations (and systems of such). From $$ax+by+cz=d$$ you can solve $x$ when $a\neq0$, $y$ when $b\neq0$ and $z$ when $c\neq0$. And, guess what, those $a,b,c$ are exactly the partial derivatives of the function here! The same with several constraints, Jacobians and the relevant minors of matrices of the coefficients of a linear system of equations. Dec 29, 2018 at 10:31