I am looking for some additional insight into this question:
Suppose that $x(s,t) = st$, and $y = y(s,t)$ with $y(1,1) = 1$ and $\frac{\partial y}{\partial t}(1,1) = 2$.
Suppose that $u = x^2 - y^2$, and $v = v(x,y)$ with $\frac{\partial v}{\partial y}(1,1) = -3$.
Now, suppose that $f(s,t) = u\bigr(x(s,t),y(s,t)\bigr)$ and $g(s,t) = v\bigr(x(s,t),y(s,t)\bigr)$, where it is known that $\frac{\partial f}{\partial s}(1,1) = -4$, $\frac{\partial f}{\partial t}(1,1) = -2$ and $\frac{\partial g}{\partial s}(1,1) = -1$.
What must be the value of $\frac{\partial g}{\partial t}(1,1)$?
I solved this simply by writing out the partial derivatives $g_t$, $g_s$, $f_t$ and $f_s$ and proceeding by substitution (see below); however, I was curious to know if there is any another method that is either more direct, faster or more revealing? perhaps using matrices (i.e. the appropriate Jacobians) to relate the two change of variables?
Here are the details of my method of solution: \begin{align} (1) \qquad g_t & = \frac{\partial v}{\partial t} = \frac{\partial v}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial v}{\partial y} \frac{\partial y}{\partial t} \qquad \qquad % (2) \qquad g_s = \frac{\partial v}{\partial s} = \frac{\partial v}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial v}{\partial y} \frac{\partial y}{\partial s} \\[5pt] % (3) \qquad f_t & = \frac{\partial u}{\partial t} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial t} \qquad \qquad % (4) \qquad f_s = \frac{\partial u}{\partial s} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial s} \\ \end{align}
From (4) we have
$$ \frac{\partial y}{\partial s} % = \frac{f_s - u_x x_s}{u_y} \Big|_{(1,1)} % = \frac{(-4) - (2x)(t)}{-2y} \Big|_{(1,1)} % = 3. $$
Plugging into (2), we can solve for $\frac{\partial v}{\partial x}$:
$$ \frac{\partial v}{\partial x} % = \frac{g_s - v_y y_s}{x_s}\Big|_{(1,1)} % = \frac{-1 - (-3)(3)}{t}\Big|_{(1,1)} % = 8. $$
Using this last result in (1) allows us to compute the value $g_t$:
$$ g_t(1,1) = (v_x x_t + v_y y_t\big|_{(1,1)} % = (8)(1) + (-3)(2) % = 2 $$
Your solution seems to be ok, I don't know others direct method.
To be more effective you can try to write down the equation in the short notation form using only the equation that are strickly needed.
For example, I've solved in this way:
$$\begin{cases} g_t=v_s= v_x \cdot x_t + v_y \cdot y_t= v_x \cdot 1 + (-3) \cdot 2=v_x-6 \\ g_s=v_s=v_x \cdot x_s + v_y \cdot y_s=v_x \cdot 1 + (-3) \cdot y_s=-1\implies v_x=3y_s-1 \\ f_s=u_x \cdot x_s + u_y \cdot y_s=2 \cdot 1 + (-2) \cdot y_s=-4\implies y_s=3 \end{cases}$$
[NOTE: you don't need to consider $f_t$]
Thus:
$$g_t=v_x-6=3y_s-7=9-7=2$$
Of course, if you had a lot of equation you could write down them all and solve the linear system obtained.