# Multivariable Chain Rule for Implicit Multivariable Functions?

I'd like to compute $$\frac{\partial x}{\partial z}$$ along $$S$$ at $$(x,y,z)$$ for $$S: \frac{1}{x}+\arctan(y+2z)=1$$.

My Approach: I can define $$w(x,y,z)=\frac{1}{x}+\arctan(y+2z)$$ and find the total differential and so on, i.e., $$dw=w_x dx+w_y dy+w_z dz$$ (we'd also need to use the fact that $$y$$ is held constant and $$dw=0$$).

How can I use the multivariable chain rule here? I'd like to find $$\frac{\partial x}{\partial z}$$ using the chain rule, but I'm a little bummed out here because I am only used to using the chain rule for solving equations where, say, $$y$$ depends on $$a,b$$ and $$a, b$$ depend on $$t$$ (e.g., $$\frac{dy}{dt}=\frac{\partial y}{\partial a}\frac{da}{dt}+\frac{\partial y}{\partial b}\frac{db}{dt}$$).

We can write \begin{align*} S:\frac{1}{x}+\arctan(y+2z)=1\tag{1} \end{align*} as function in $$x=x(y,z)$$.
We obtain from (1) \begin{align*} x(y,z)&=\frac{1}{1-\arctan(y+2z)}\\ \\ \color{blue}{\frac{\partial x}{\partial z}(z,y)} &=\frac{\partial}{\partial z}\left(\frac{1}{1-\arctan(y+2z)}\right)\\ &=\frac{\frac{\partial}{\partial z}\left(\arctan(y+2z)\right)}{\left(1-\arctan(y+2z)\right)^2}\tag{2}\\ &=\frac{2}{\left(1+(y+2z)^2\right)\left(1-\arctan(y+2z)\right)^2}\\ &\,\,\color{blue}{=\frac{2x^2}{1+(y+2z)^2}}\tag{3} \end{align*}
• In (2) we use $$\left(\frac{1}{g(z)}\right)^{\prime}=-\frac{\left(g(z)\right)^{\prime}}{(g(z))^2}$$.
• In (3) we use $$x=\frac{1}{1-\arctan(y+2z)}$$.