$f\left(x,y\right) = x^4 + 3y^{\frac{4}{3}}$ with restriction $xy = c$ and the constant $c > 0$. I can't find a way to find the critical points with Lagrange multipliers.
I put together the following system of equations:
$$ L_x = 4x^3 - y \cdot \lambda = 0 \\ L_y = 4\sqrt[3]{y} - x \cdot \lambda = 0 \\ L_\lambda = x \cdot y - c=0 $$
On
To avoid the multiplier $\lambda$, in this kind of cases ($n=2$ with only one constraint) it is recommended to work with a determinant as follows to reduce the dimension of the system obtained by the LM.
Let $g(x, y)=xy-c$ be the constraint function ($xy=c$ is equivalent to $g(x, y)=0$). Since for your critical point $\nabla f(x, y)=\lambda\nabla g(x, y)$, then the vectors $\nabla f$ and $\nabla g$ are linearly dependent, so \begin{align*} \text{det}(\nabla f\ \ \nabla g)=0\ \Longrightarrow 4x^4-4y^{4/3}=0. \end{align*} It follows that $x=y^{1/3}$ and then from the constraint $xy=c>0$ you can obtain the critical point(s). Here there is only one and by inspection it is a maximum.
There is a critical point at $(0,0).$ As for the others.
From the 3rd equation: $x = \frac {c}{y}$
Substituting into the first equation:
$4x^3 - y\lambda = 0\\ \frac {4c^3}{y^3} = y\lambda\\ \lambda = 4c^3y^4$
And the 2nd:
$4y^{\frac 13} = \lambda x\\ 4y^{\frac 13} = 4c^3y^4\frac {c}{y}\\ y^{-\frac {8}{3}} = c^4\\ y = 2^{-\frac {3}{4}} c^{-\frac 32}$
knowing $y$ we can find $x$ and $f(x,y).$
$x = \frac {c}{y} = 2^{\frac 34}c^{\frac 52}\\ f(x,y) = x^4 + 3y^{\frac 43} =8c^{10}+\frac {3}{2c^2}$
And there should also be a maximum at $(x,y) = (-2^{\frac 34}c^{\frac 52},-2^{-\frac 34} c^{-\frac 32})$