I'm trying to attempt the following exercise:
Show that the set of all points $(x, y) \in \mathbb{R}^2$ such that $(x + y)^5 - xy - 1 = 0$ is a $1-$manifold.
Here's my attempt:
Let $F(x,y) : \mathbb{R}^2 \longrightarrow \mathbb{R}$ be defined by $F(x, y) = (x + y)^5 - xy - 1$. Clearly, $F(x,y)$ is a $\mathcal{C}^1$ function since it is a polynomial function in two variables. Now, we have that if $F_{x}(x,y) = 0$ or $F_{y}(x,y) = 0$ for each point $(x,y)$ on the level set in $\mathbb{R}^2$, then the conclusion follows easily from the Implicit Function Theorem.
Now, we have that both the partial derivatives are zero if and only if
\begin{align} F_{x}(x,y) = 5(x + y)^4 - y = 0 = 5(x + y)^4 - x = F_{y}(x,y) \end{align}
which implies that $x = y$. Let's see if the function $F(x,y)$ is identically zero on the line $y = x$ in $\mathbb{R}^2$. Let,
\begin{align} G(x) = F(x,x) = (x + x)^5 - x^2 - 1 = 32x^5 - x^2 - 1. \end{align}
Now $G(0) = -1$ and $G(1) = 30$ imples that $G(x)$ has a root in $(0,1)$. It can be easily checked that $G(x)$ is a seperable polynomial, so there's only one real root.
So, I have found a number $c \in (0,1)$ such that,
\begin{align} F(c,c) = 0, \quad \nabla F(c,c) = 0. \end{align}
At all other points, it is clear that $F(x,y)$ defines a $1-$manifold. What about about the point $c$ given above. How do I analyze this point?
If $x=y$ holds it doesn't mean that the gradient would be zero. As noted from above for the gradient to be zero we must have $x=y$, not the other way around. Nevertheless substitute $x=y$ in the partial derivatives and then we want to solve:
$$0 = 5\cdot 16x^4 - x = 80x^4 - x$$
Additionally $F(x,x) = 0$ must be satisfied which gives us that $32x^5 - x^2 - 1 = 0$. Substitute from above to get:
$$0 = 32x^5 - x^2 - 1 = 32x^5 - 80x^5 - 1 = -48x^5 - 1 \implies x = \frac{1}{-\sqrt[5]{48}}$$
However it's not hard to notice that for this point we have that $80x^4 - x=0$ isn't satisfied. Hence for all points in the inverse image the gradient isn't $0$ and so they constitute a $1-$manifold.