Say I have a function $f(x,y,z)$. If I know $t= \sqrt{x+\sqrt{x^2+ y*z}}$ and I know the partials $\large\frac{\partial{f}}{\partial{x}}$,$\large\frac{\partial{f}}{\partial{y}}$,$\large\frac{\partial{f}}{\partial{z}}$, how could I apply the chain rule in order to obtain $\large\frac{\partial{f}}{\partial{t}}$?
I would have thought:
$$\frac{\partial{f}}{\partial{t}} = \frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{t}} + \frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{t}} +\frac{\partial{f}}{\partial{z}}\frac{\partial{z}}{\partial{t}}$$
however, I already feel like i'm on the wrong track. Can anyone give me a start on how to construct $\large\frac{\partial{f}}{\partial{t}}$ in terms of $\large\frac{\partial{f}}{\partial{x}}$,$\large\frac{\partial{f}}{\partial{y}}$ and $\large\frac{\partial{f}}{\partial{z}}$?
Your deduction is formally correct i.e. it is true that $$ \frac{\partial{f}}{\partial{t}} = \frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{t}} + \frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{t}} +\frac{\partial{f}}{\partial{z}}\frac{\partial{z}}{\partial{t}}\label{1}\tag{1} $$ However, you missed a basic step: with your definition of $t$, how can you deduce three functions $$ t\mapsto \big(x(t), y(t), z(t)\big)\; $$ i.e. define a function $\boldsymbol{p}:\mathbb{R}\to \mathbb{R}^3$ such that $f(t)\triangleq f(\,\boldsymbol{p}(t))=f\big(x(t), y(t), z(t)\big)$?
The relation you are considering, i. e. $$ t= \sqrt{x+\sqrt{x^2+ y\cdot z}}\label{2}\tag{2} $$ defines a 2-dimensional variety in $\mathbb{R}^3$, and in order to parametrize it two parameters are needed, therefore you cannot use \eqref{2} to define any such $\boldsymbol{p}=\boldsymbol{p}(t)$.