= Solution
A <function> $f:D\to\mathbb R$ on a <convex set> $D$ is <convex> when
$$
f((1-t)x+ty)
\leq(1-t)f(x)+tf(y)
$$
for all $x,y\in D$ and $0\leq t\leq1$.
If $f$ is once <differentiable>, this is equivalent to monotonicity of its <gradient>:
$$
\boxed{
(\nabla f(x)-\nabla f(y))\cdot(x-y)\geq0
}.
$$
Equivalently, every tangent hyperplane supports the graph:
$$
f(y)\geq f(x)+\nabla f(x)\cdot(y-x).
$$
If $f$ is twice <differentiable>, convexity is equivalent to the <Hessian matrix> being positive semidefinite throughout $D$.
For
$$
f(x,y)=x^3+y^3+Axy,
$$
the Hessian is
$$
H=
\begin{pmatrix}6x&A\\A&6y\end{pmatrix}.
$$
A real symmetric two-by-two <matrix> is positive semidefinite exactly when its two diagonal entries and its <determinant> are nonnegative. Thus
$$
x\geq0,\qquad y\geq0,\qquad 36xy-A^2\geq0.
$$
The largest convexity domain is therefore
$$
\boxed{
D_A=\{(x,y):x\geq0,\ y\geq0,\ 36xy\geq A^2\}
}.
$$
For $A\ne0$, its boundary is the hyperbola $y=A^2/(36x)$ in the first quadrant and the domain lies above it. For $A=0$, it is the closed first quadrant. This is the <convexity domain of x cubed plus y cubed plus Axy>.
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