Are the following functions convex/concave? $f(x, y) = x^{3}-\ln(x^2y)$ and $f(x,y)=x^\alpha y^\beta-x-y$, where both $\alpha, \beta\in(0,1)$ The definition of concave/convex function in a open,convex set A is:

Let $f : A ⊆ R^n −→ R$ with $A$ open and convex and let $f ∈ C^2(A)$;
  then $f$ is convex in $A$ if and only if, for all $x ∈ A$, $D2f(x)$
  (Hessian) is Positive SD;  $f$ is concave in A if and only if, for all
  $x ∈ A, D2f(x)$ is Negative SD.

I would like to know whether the following functions are convex/concave in their maximal domain of existence. If not, is there a 'maximal' open set A where they are convex/concave?
1) $f(x, y) = x^{3}-\ln(x^2y)$.
2) $f(x,y)=x^\alpha y^\beta-x-y$, where both $\alpha, \beta \in(0,1)$.
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Here is my attempt:

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1)
The maximal interval is:
$D=\{(x, y) ∈ R^2 : x \neq 0, y>0\}$
Clearly this set is not convex, so that we can not have convexity/concavity of f. Looking at the Hessian:
$H=\pmatrix{6x+\frac{2}{x^2}&0 \cr 0&\frac{1}{x^2}\cr}$
We get that the function is convex in $A=\{(x, y) ∈ R^2 : x > 0, y>0\}$
2)
The maximal interval is:
$D=\{(x, y) ∈ R^2 : x \geq 0, y\geq 0\}$
Clearly this set is not open, so that we can not have convexity/concavity of f. But I think we can have convexity/concavity in $A=\{(x, y) ∈ R^2 : x > 0, y>0\}$  Looking at the Hessian:
$H=\pmatrix{\alpha(\alpha-1)x^{\alpha-2}y^\beta&\beta(\alpha)x^{\alpha-1}y^{\beta-1}\cr \beta(\alpha)x^{\alpha-1}y^{\beta-1}&\beta(\beta-1)x^\alpha y^{\beta-2}\cr}$
We get then that the determinant is: $\beta (\beta-1)\alpha (\alpha-1) x^{2(\alpha-1)}y^{2(\beta -1)}-\beta^2 \alpha ^2 x^{2(\alpha-1)}y^{2(\beta -1)}$, where the term $a_{11}$ of the hessian matrix is $\alpha(\alpha-1)x^{\alpha-2}y^\beta\leq0$ $\forall x,y$. So the hessian is NSD if the determinant is $\geq 0$. But it looks like the sign of the determinant depends on $\alpha,\beta$. But I can not get for which values of $\alpha, \beta$ the determinant is positive or zero.
Can you help me? Is my first attempt correct? And also what about the second one??
 A: Your answer is correct. The first function is convex in the region you found and there is no way for the first function to be concave because the determinant and the first principal minor in the Hessian of the first function always have the same sign. For the second part though, we need the determinant and the first principal minor($a_{11}$) to be positive  thus $\beta(\beta-1)\alpha(\alpha-1)x^{2(\alpha-1)}y^{2(\beta -1)}-\beta^2\alpha^2x^{2(\alpha-1)}y^{2(\beta -1)}\ge0\Rightarrow\beta(\beta-1)\alpha(\alpha-1)-\beta^2\alpha^2\ge0\Rightarrow\alpha\beta(1-\alpha-\beta)\ge0$.
Also $\alpha(\alpha-1)x^{\alpha-2}y^\beta \ge 0 \Rightarrow \alpha(\alpha-1)\ge 0$ because the domain is $x\ge0,y\ge0$. Thus we need to find where these two expression $f(\alpha ,\beta)=\alpha\beta(1-\alpha-\beta),g(\alpha ,\beta)=\alpha(\alpha-1)$ are positive. In the following figure we see where $f(\alpha,\beta)$ is positive. Four regions exist $\alpha,\beta \in [0,1]$ or $\alpha\ge1,\beta\le0$ or $\alpha\le0,\beta\ge1$ or $\alpha\le0,\beta\le0$. We can find these regions from the fact that the multiplications of three factor $\alpha$, $\beta$ and $(1-\alpha-\beta) $ must be positive and since each factor is linear, there are $2^3-1$ regions total (rach one being positive or negative)(please consider the horizontal axis as $\alpha$ and the vertical axis as $\beta$)

In the second figure we see where the second one $g(\alpha,\beta)$ is positive. This one is easier, since is only a function of $\alpha$ and we have two linear factors and $2^2-1$ regions total.

and by intersection we arrive at the three regions that satisfies convexity $\alpha\ge1,\beta\le0$ or $\alpha\le0,\beta\ge1$ or $\alpha\le0,\beta\le0$. There is also the assumption that $\alpha,\beta \in (0,1)$. If we take this into consideration, then there is no way this function can be convex. But hopefully there is one region that this function can be concave. For concavity the Hessian must be negative definite or equivalently, $a_{11}$ must be negative and the determinant must be positive, Therefore $f(\alpha,\beta)$ must be positive and $g(\alpha,\beta)$ must be negative. The only region satisfying this according to above explanations is the intersection of the region $\alpha,\beta \in (0,1)$ and under the line $\alpha+\beta=1$, the triangular region $ABC$ in the first figure.
