Problem
Let $M$ be the set of all points $(x,y) \in \mathbb{R}^2$ satisfying the equation
$$xy^3 + \frac{x^4}{4} + \frac{y^4}{4} = 1 $$
Prove that $M$ is a manifold. What is the dimension of $M$?
Attempt
I think this question it started by saying the following:
$$\phi=xy^3 + \frac{x^4}{4} + \frac{y^4}{4} - 1$$
Not overly sure how do this question so any help in the right direction would be appreciated. Anyway, I got the partial derivatives:
$$\frac{{\partial}\phi}{{\partial}x}=y^3 + x^3$$
$$\frac{{\partial}\phi}{{\partial}y}=3xy^2 + y^3$$
After here I'm stuck, I can't find any clear way of answering this. Thanks in advance for any help.
You should prove that $0$ is regular value of function $\phi: \mathbb{R}^2 \to \mathbb{R}$, that means in practise that matrix:
$$\begin{bmatrix} y^3 + x^3 & 3xy^2 + y^3 \end{bmatrix}$$
is not zero matrix if $\phi(x,y)=0$.
$\textbf{Edit:}$(How to prove that zero is regular value of function $\phi$). Suppose that for $(x,y)$:
$$\begin{bmatrix} y^3 + x^3 & 3xy^2 + y^3 \end{bmatrix}=\begin{bmatrix}0 & 0\end{bmatrix}$$
Then $x=-y$ (by first coordinate). Next let's look on second coordinate. If $x=-y$ then $-3y^2+y^3=2y^3=0$, so $x=y=0$. So $(x,y)=(0,0)$. But we check that if $(x,y)=(0,0)$ then $\phi(x,y) \neq 0$.
This way you show that $M$ is submanifold of $\mathbb{R}^2$, but every submanifold is manifold.
Dimension of $M$ is equal to $\dim \mathbb{R}^2 -\dim \text{Im}D\phi=2-1=1$.