I have to find the values of $\lambda$ (real) for which $f(x, y) = -(x-y)^4 + \lambda x^{2/3}$ is concave.
Attempts
I know that $(x-y)^4$ is convex in $\mathbb{R}^2$ hence $-(x-y)^4$ is concave. Since sum of concave functions is concave, then I must find $\lambda$ for which $x^{2/3}$ is concave.
I know that in $\mathbb{R}$, $x^{2/3}$ is concave, hence it's concave in $\mathbb{R}^{2+}$ too.
So I would say that for $\lambda \geq 0$, $f(x, y)$ is concave in $\mathbb{R}^+$.
Yet I tried to plot the whole function for $\lambda = -2$ and it looks like concave too...
What am I missing?
ADD
Can I say that, studying the function alont $y \to 0$, I can reduce the problem to the study of
$$f(x, 0) = -x^4 + \lambda x^{2/3}$$
and by observing that for $\lambda < 0$, the function is convex, hence not concave in $\mathbb{R}^{2+}$?
We know that a function f, supposed differentiable, is concave if and only if its hessian matrix $H_{f}(x,y)$ is negative semi-definite for all $(x,y)\in\mathbb{R}^{2+}$. Deriving, we get that $$ H_{f}(x,y)= \begin{pmatrix} -\frac{2}{9}\lambda x^{-4/3}-12(x-y)^2 & 12(x-y)^2 \\ 12(x-y)^2 & -12(x-y)^2 \end{pmatrix} $$ And then what's left is to determine $\lambda$ such that this matrix is negative semi-definite.