Chain rule when $u$ is a fucntion of $x$ and $y$ and both $x$ , $y$ are fucntions of $u$ and $v$

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Find $\frac{\partial u}{\partial x}$ where $u$ and $v$ are functions of $x$ and $y$ $\bigg($ $u(x,y)$ and $v(x,y)$ $\bigg)$ and $x$ and $y$ is of the form, $x = f(u,v), y=g(u,v)$.

This was in an exam and had multiple choices for answers. Some answers contained $\partial f$ and $\partial g$ terms.

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Use implicit differentiation on $x=f(u,v)$, considering $u$ and $v$ as functions of $x$ and $y$, to obtain (differentiation w.r.t. $x$): $$1 = \frac{\partial f}{\partial u}\color{blue}{\frac{\partial u}{\partial x}} + \frac{\partial f}{\partial v}\color{red}{\frac{\partial v}{\partial x}}$$ Doing the same (differentiation w.r.t. $x$) for $y=g(u,v)$ gives: $$0 = \frac{\partial g}{\partial u}\color{blue}{\frac{\partial u}{\partial x}} + \frac{\partial g}{\partial v}\color{red}{\frac{\partial v}{\partial x}}$$ Together this gives you a system of two linear equations in the unknowns $\color{blue}{u_x}$ and $\color{red}{v_x}$; you can solve for $u_x$ and compare with the multiple choice alternatives.

Addendum: if you use implicit differentiation w.r.t. $y$ as well, you can also find $u_y$ and $v_y$. You may recognize a common denominator appearing in all these partial derivatives: this is the Jacobian of the transformation between both pairs of variables.