Compute $F_\ast(\partial/\partial x)$ as a linear combination of $\partial/\partial u, \partial/\partial v$ and $\partial/\partial w$. 
Let $F: \Bbb R^2 \to \Bbb R^3$ be the map $(u,v,w)=F(x,y)=(x,y,xy).$ Compute $F_\ast(\partial/\partial x)$ as a linear combination of $\partial/\partial u, \partial/\partial v$ and $\partial/\partial w$.

So if $F_\ast$is the differential then it's defined as $F_\ast : T_p\Bbb R^2 \to T_{F(p)}\Bbb R^3$. The vectors $\partial/\partial u, \partial/\partial v$ and $\partial/\partial w$ serve as a basis for $T_{F(p)}\Bbb R^3$ so we need to determine the constants $a,b,c$ from $$F_\ast(\partial/\partial x)=a \frac{\partial}{\partial u} + b \frac{\partial}{\partial v} + c \frac{\partial}{\partial w}?$$
I don't quite understand the setting here. The differential $F_\ast(\partial/\partial x)$ means nothing if it's not acting on some function so what does this equation even represent?
 A: Differentials of maps $F:\mathbb{R}^n \to \mathbb{R}^m$ are just the total derivative. On general manifolds, we take the total derivative of the coordinate representation of $F$. I suggest brushing up on the total derivative in $\mathbb{R}^n$, even the case of $\mathbb{R}^n \to \mathbb{R}$ might help, see here. It is the directional derivative in all directions, given in terms of partial derivatives--the change in each coordinate direction.
The formula for the "differential" on a coordinate basis vector with respect to some charts is basically the chain rule (more details are given in any differential geometry text). In this case you will in fact have what Vajra wrote, but did not simplify:
$F_{*} \left( \frac{\partial}{\partial x} \right) =\frac{\partial u}{\partial x}\frac{\partial}{\partial u}+\frac{\partial v}{\partial x}\frac{\partial}{\partial v}+\frac{\partial w}{\partial x}\frac{\partial}{\partial w}=\frac{\partial}{\partial u}+y\frac{\partial}{\partial w}.$
This is your answer. If you do the same for $F_{*} \left( \frac{\partial}{\partial y} \right)$, you can find that the matrix of $F_{*}$ with respect to the bases $\{\frac{\partial}{\partial x}|_p,\frac{\partial}{\partial y}|_p\}$ for $T_p\mathbb{R}^2$ and $\{\frac{\partial}{\partial u}|_p,\frac{\partial}{\partial v}|_p,\frac{\partial}{\partial w}|_p\}$ for $T_p\mathbb{R}^3$ is the 3x2 matrix $$\begin{pmatrix}
1 &0 \\
0 &1 \\
y &x 
\end{pmatrix}.$$
This is none other than the Jacobian of $F$.
A: $\mathbb R^2$ has a global system of coordinates $\{x,y\}$ and the function $f:\mathbb R^2\to\mathbb R^3$ defined as $$f(x,y)=\begin{pmatrix}u\\v\\w \end{pmatrix}=\begin{pmatrix}x\\y\\xy \end{pmatrix}\text{ is }\mathcal C^{\infty}(\mathbb R^3).$$In  a given point $p\in\mathbb R^2$, $f$ induces the pushforward map $f_{*p}:T_p\mathbb R^2\to T_{f(p)}\mathbb R^3$.
The pushforward is linear (remember that $T_{p}\mathbb R^2\cong\mathbb R^2$and $T_{f(p)}\mathbb R^3\cong \mathbb R^3$) and the standard coordinates of $\mathbb R^2$ define a basis for $T_p\mathbb R^2$, which is $\mathcal A=\{\partial_x\vert_p,\partial y\vert_p,\partial z\vert_p\}$. At the same way, the coordinates $\{u,v,w\}$ in $\mathbb R^3$ define the basis $\mathcal B=\{\partial u\vert_{f(p)},\partial v\vert_{f(p)},\partial w\vert_{f(p)}\}$ for $T_{f(p)}\mathbb R^3$. The jacobian of $f$ evaluated in $p$ is the representative matrix of $f_{*p}$ respect to the bases $\mathcal A$ and $\mathcal B$.
If you want to compute the imagine of the basis of $T_p\mathbb R^2$ through $f_{*p}$ you can use the matrix associated to the linear map $f_{*p}$ (so the matrix $\nabla f\vert_p$). More directly, $$\begin{align}
f_{*p}\left(\frac{\partial}{\partial x}\bigg\vert_p\right)=\frac{\partial u}{\partial x}\bigg\vert_p\frac{\partial }{\partial u}\bigg\vert_{f(p)}+\frac{\partial v}{\partial x}\bigg\vert_p\frac{\partial }{\partial v}\bigg\vert_{f(p)}+\frac{\partial w}{\partial x}\bigg\vert_p\frac{\partial }{\partial w}\bigg\vert_{f(p)}\\
\end{align}.$$
Same thing for the other two coordinates.
