Does this entangled PDE capture the derivative? In a game-theory textbook, I encountered the following. Suppose we have player $1$ and and $2$ optimizing by playing strategies $x_1$ and $x_2$ $\in \mathbb{R}$. The first order conditions for player $1$ and $2$ are given respectively by:
$$f(x_1,x_2,a,b)=1$$
$$g(x_1,x_2,a,b)=1$$
What is $\frac{\partial x_1}{\partial b}$? Naively, we may apply the implicit function theorem:
$$\frac{\partial x_1}{\partial b}=-\frac{\frac{\partial f(..)}{\partial a}}{\frac{\partial f(..)}{\partial x_1}}$$
But given that $a$ is also a function of the first order condition for the second player, we must take into account the effect of $a$ through its effect on $x_2$. Hence, the PDE is given by
$$\frac{\partial x_1}{\partial a}=-\frac{\frac{\partial f(..)}{\partial a}+\frac{\partial x_2}{\partial a}*\frac{\partial f(..)}{\partial x_2}}{\frac{\partial f(..)}{\partial x_1}+\frac{\partial x_2}{\partial x_1}*\frac{\partial f(..)}{\partial x_2}}$$
Where $x_2$ satisfies player $2$'s FOC. In the numerator, the term $\frac{\partial x_2}{\partial a}*\frac{\partial f(..)}{\partial x_2}$ captures the effect of $a$ through $x_2$ on $f(..)$, and the $\frac{\partial x_2}{\partial x_1}*\frac{\partial f(..)}{\partial x_2}$ the effect of $x_1$ through $x_2$ on $f(..)$. Is this a correct application of the inverse function theorem that produces a good approximation of $\frac{\partial x_1}{\partial b}$?

One factor that makes me hesitant about this PDE is in the numerator, the term $\frac{\partial x_2}{\partial a}*\frac{\partial f(..)}{\partial x_2}$ captures the effect of $a$ on $x_2$. However, this is exactly what we're trying to define all along, albeit for $x_1$. At any rate, this part will not completely characterize the effect of $a$ on $x_2$. 

If you differentiate both equations in $a$, you get
$$
f_a+x^1_af_1+x^2_af_2=0,\\
g_a+x^1_ag_1+x^2_ag_2=0,
$$
where I'm denoting $x^j_a=\partial x^j/\partial a$ and $f_j=\partial f/\partial x^j$.  If we solve this system for $x^1_a$, we get
$$
x^1_a=\frac{f_2g_a-g_2f_a}{g_2f_1-g_1f_2}.
$$
Using Jacobians, you can express this as
$$
\frac{\partial x^1}{\partial a}=-\frac{\partial(f,g)/\partial(a,x^2)}{\partial (f,g)/\partial(x^1,x^2)}.
$$
