Finding the range of a vector valued function For a single valued function, I can infer if the function is monotone from its derivative.
For a vector valued function, is it possible to infer monotonicity from the directional derivative?
For example, define 
$$
D=[1,2]\times[1,2],
$$ 
and
$$
f(x,y)=\left( \frac{2}{1/x+1/y},\sqrt{xy} \right).
$$
Is it possible to show that $f$ maps $D$ to $D$ from its gradient $\nabla f$?
The gradient is
$$
\nabla f = \begin{pmatrix}
\frac{2}{\left(1+x/y\right)^2} & \frac{2}{\left(1+y/x\right)^2} \\
\frac{y^{1/2}}{2x^{1/2}} & \frac{y^{1/2}}{2x^{1/2}}
\end{pmatrix},
$$
whence, for $(x,y)$ in $D$, the directional derivative
$$
\left(\nabla f(x,y)\right)\begin{pmatrix} x \\ y \end{pmatrix},
$$
is always positive and I would like to conclude that, on $D$,
$$
\text{$f$ is minimal at $(1,1)$},\\
\text{$f$ is maximal at $(2,2)$}.
$$
Is it the right way to proceed?
The graph of each component of $f$ looks like this

 A: I'm not quite sure what you mean by monotonicity for a function of several variables, let alone a vector-valued function. My approach would be to start by seeing where $f$ maps the boundary of $D$. Note that, by the symmetry, it covers the curve 
$$\left\{\left(\frac{2x}{x+1},\sqrt x\right): 1\le x\le 2\right\} \cup \left\{\left(\frac{4x}{x+2},\sqrt{2x}\right): 1\le x\le 2\right\}$$
twice. The rest of the boundary of the image comes from the diagonal:
$$\left\{(x,x): 1\le x\le 2\right\} \,.$$
A Mathematica image of the boundary of $f(D)$ is this:

Some thought about the Inverse Function Theorem will tell you that the mapping is open away from critical points (points where $Df$ is singular).
A: Note that the Jacobian determinant is
$$\det J_f(x,\, y)=\det\begin{bmatrix}
\frac{2}{\left(1+x/y\right)^2} & \frac{2}{\left(1+y/x\right)^2} \\
\frac{y^{1/2}}{2x^{1/2}} & \frac{y^{1/2}}{2x^{1/2}}
\end{bmatrix}=\\
=\dfrac{1}{{\left(\dfrac{1}{x} + \dfrac{1}{y}\right)}^{2} \sqrt{x y} x} - \dfrac{1}{{\left(\dfrac{1}{x} + \dfrac{1}{y}\right)}^{2} \sqrt{x y} y}= \\
=\dfrac{y-x }{{\left(\dfrac{1}{x} + \dfrac{1}{y}\right)}^{2}  (xy)^{\frac{3}{2}}},$$ therefore $\det J_f(x,\, y)=0$ on the line $y=x.$
Addition:
Denote $$G_1=\{(x, \ y)\colon \;\; {1} < {x} < {2},\;\; x < {y} < {2} \}, \\
G_2=\{(x, \ y)\colon \;\; {1} < {y} < {2},\;\; y < {x} < {2} \}$$ Then $J_f (x,\ y)\ne{0}, \;\;\; \forall(x, \ y)\in{G_1\cup G_2},$
thus $f(x, \ y)$ is a diffeomorphism on each $G_1$ and $G_2$  and
$$f(\partial{G_1})=\partial{f(G_1 )}, \\
f(\partial{G_2})=\partial{f(G_2 )}.$$
As noted by Ted Shifrin, $$\gamma_1=\left\{ f(1,\ y),\;\;  1 < y < 2\right\}=\left\{\left(\dfrac{2y}{y+1},\sqrt y\right)\colon \;\;  1 < y < 2\right\}, \\
\gamma_2=\left\{ f(x,\ 2),\;\;  1 < x < 2\right\} = \left\{\left(\frac{4x}{x+2},\sqrt{2x}\right)\colon \;\;  1 < x < 2\right\}, \\
\gamma_3=\left\{ f(x,\ x),\;\;  1 < x < 2\right\} = \left\{\left(x,x\right)\colon \;\; 1 < x < 2\right\}.$$
Therefore the interior of $\partial{G_1}$ is mapped by $f$ into the interior of $f(\partial{G_1}) = \gamma_1\cup\gamma_2\cup \gamma_3.$
Due to the symmetry of  $f$ with respect to $x$ and $y$ the interior of $\partial{G_2}$ is mapped by $f$ also into the interior of $\gamma_1\cup\gamma_2\cup \gamma_3.$
Different signs of the Jacobian determinant on both sides of the diagonal indicate the opposite orientation of parts of the image of $D.$ 
