We have $$x = x(u,v) \ \ \ \text{and} \ \ \ y = y(u,v)$$
then the chain rule for functions of several variables states that
$$ \frac{\partial x}{\partial x} = \frac{\partial x}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial x}$$ But $$\frac{\partial x}{\partial x} = 1$$
I'll use these facts later on. For now,
I believe we are trying to prove $$ \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = I$$ where $$I = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$
We'll also prove $$\Bigg \lvert \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)}\Bigg \rvert = 1$$
So we begin:
$$\frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix}\begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{vmatrix}$$
$$ = \det \Bigg (\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix} \Bigg ) \det \Bigg (\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{bmatrix} \Bigg )$$
$$ = \det \Bigg ( \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix}\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{bmatrix} \Bigg )$$
by the identity $$\det(AB) = \det(A) \det(B)$$
so $$ \det \Bigg ( \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix}\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{bmatrix} \Bigg )$$
$$ = \det \Bigg ( \begin{bmatrix} \frac{\partial x}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial x} & \frac{\partial x}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial y} \\ \frac{\partial y}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial y}{\partial v} \frac{\partial v}{\partial x} & \frac{\partial y}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial y}{\partial v} \frac{\partial v}{\partial y}\end{bmatrix} \Bigg )$$
$$ = \det \Bigg ( \begin{bmatrix} \frac{\partial x}{\partial x} & \frac{\partial x}{\partial y} \\ \frac{\partial y}{\partial x} & \frac{\partial x}{\partial x}\end{bmatrix} \Bigg )$$
$$ = \det \Bigg ( \begin{bmatrix} 1 & \frac{\partial x}{\partial y} \\ \frac{\partial y}{\partial x} & 1\end{bmatrix} \Bigg ) $$
by the chain rule for functions of several variables, using what we found above.
Now, since $x$ is not a function of $y$, and $y$ is not a function of $x$, we have $$ \frac{\partial x}{\partial y} = 0 \ \ \ \text{and} \ \ \ \frac{\partial y}{\partial x} = 0$$
so $$ \det \Bigg ( \begin{bmatrix} 1 & \frac{\partial x}{\partial y} \\ \frac{\partial y}{\partial x} & 1\end{bmatrix} \Bigg ) = \det \Bigg (\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} \Bigg ) = \det(I)$$
so we've proven that $$ \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = I $$
Now, $$ \det(I) = \det \Bigg (\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} \Bigg ) = 1 $$
so we've also proven that $$ \Bigg \lvert \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)}\Bigg \rvert = 1 $$ as required
You can check this by going back to the definitions and properties of linear transformations.
Suppose $A:\mathbb R^2\to \mathbb R^2$ is a linear transformation such that $A((1,0))=a(1,0)+b(0,1)$ and $A((0,1))=c(1,0)+d(0,1)$. This determines $A$ uniquely and the matrix with respect to the basis $(1,0),(0,1)$ is
\begin{bmatrix}
a& c\\
b & d
\end{bmatrix}
Now,
$A((1,0))=a(1,0)+b(0,1)=a(1,0)+b((1,1)-(1,0))=(a-b)(1,0)+b(1,1)$
and
$A((1,1))=A((1,0))+A((0,1))=(a+c)(1,0)+(b+d)(0,1)=(a+c)(1,0)+(b+d)((1,1)-(1,0))=((a+c)-(b+d))(1,0)+(b+d)(1,1)$
Then, the matrix with respect to the basis $(1,0), (1,1)$ is
\begin{bmatrix}
a-b& (a+c)-(b+d)\\
b & b+d
\end{bmatrix}
Best Answer
It is a little awkward, but using the partial derivative notation with vectors basically means:
$$\begin{align}\dfrac{\partial (a, b, c)}{\partial (x, y, z)} ~=~& \det\left((\dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z})^\top(a,b,c) \right)^\top \\[1ex] =~& \begin{vmatrix}\dfrac{\partial a}{\partial x}&\dfrac{\partial a}{\partial y}&\dfrac{\partial a}{\partial z}\\ \dfrac{\partial b}{\partial x}&\dfrac{\partial b}{\partial y}&\dfrac{\partial b}{\partial z}\\\dfrac{\partial c}{\partial x}&\dfrac{\partial c}{\partial y}&\dfrac{\partial c}{\partial z}\end{vmatrix} \end{align}$$
The notation summarises the essentials of the Jacobian determinant. You are taking the partial derivatives of $a, b, c$ each with respect to $x,y,z$, constructing a matrix of the result and evaluating its determinant.
$$\dfrac{\partial (a, b, c)}{\partial (x, y, z)}$$