This is one of the questions we were asked in our iniversity examinations.
Find $\left(\frac{\partial u}{\partial x}\right)_y$ for the system: \begin{align} \begin{cases} x+y^3+u^3+v^3=0\\ x^3+y-u^4-v^4=0 \end{cases} \end{align} Looking at the system we can say there will be two dependent variables. Since $x$ and $y$ are independent variables as specified in the question, $u$ and $v$ will be considered as dependent variables.
I have figured out two ways through which the problem can be approached.
a) Eliminate $v$. Partially differentiate the expression containing $u$, $x$ and $y$ w.r.t $x$ treating $y$ as constant and get the answer.
b) Partially differentiate both the equations w.r.t $x$ treating $y$ as constant. Now eliminate $\left(\frac{\partial v}{\partial x}\right)_y$ and get the required answer.
What i am not able to figure out is how to draw the dependency diagram (tree diagram) and write the chain rule for the given system. Kindly help by generalizing as to how to write the formula of chain rule for a system of implicit equations.
Having system $$\begin{cases} x+y^3+u^3+v^3 = 0 & \\ x^3+y-u^4-v^4 = 0 \end{cases}$$ and differentiate each equation by $x$, assuming $u=u(x,y)$ and $v=v(x,y)$, we obtain $$\begin{cases} 1+3u^2u'_x+3v^2v'_x = 0 & \\ 3x^2-4u^3u'_x-4v^3v'_x = 0 \end{cases}$$ Now we have system of two linear equations with respect to variables $u'_x$ and $v'_x$, which can be solved by well known formulas in appropriate conditions.
For general case let's assume, that we have mapping $F$ continuous in neighborhood $W$ of the point $(x_0, y_0)\in \mathbb{R}^m\times \mathbb{R}^n$, with values in $\mathbb{R}^m$ and with continuous partial derivative $F'_y(x,y)$ in it. If $F(x_0, y_0)=0$ and there exists $(F'_y(x_0,y_0))^{-1}$ and $|| (F'_y(x_0,y_0))^{-1}||<\infty$, then there exists neighborhood $U=U(x_0)$ and neighborhood $V=V(y_0)$ and function $f\colon U \to V$, continuous at $x_0$, such that $U \times V \subset W$ and $(F(x,y)=0 \text{ within }U \times V) \Leftrightarrow (y=f(x),x \in U)$. If in addition we know, that $F$ is differentiable at the point $(x_0,y_0)$, then the implicit function is also differentiable at the point $x_0$ and holds $$f'(x_0)=-(F'_y(x_0,y_0))^{-1}F'_x(x_0,y_0)$$ More details and proofs you can find, for example, in Vladimir A. Zorich - Mathematical Analysis I-Springer, 2016, pages 595-597, or on Implicit function theorem in section "Statement of the theorem" in formula for $\frac{\partial g}{\partial x_j}$.