Given $(*)x^2 + y^2 +z^2 - 3xyz = 0$
and a fucntion $f(x,y,z) = xy^{2}z^{3}$
first section wants to show that equation $(*)$ defines a function $z = z(x,y)$ around point $(1,1,1)$ then find $f'_{x}(1,1,1)$ so it's not problem:
we create a function $g(x,y,z) = x^2 + y^2 +z^2 - 3xyz$ and we show:
(1) $ g(1,1,1) = 0 $
(2) partial derivatives of $g(x,y,z)$ are elementary functions
(3)$g'_{z} = -1 \neq 0$
so $g(x,y,z)$ defines a function $z = z(x,y)$ around point $(1,1,1)$
$f'_{x} = y^{2} z^{3} + xy^{2} \cdot3z^{2}z'_{x} |_{(1,1,1)} = 1 + 3z'_{x}$
while $z'_{x} = -\frac{g'_{x}(1,1,1)}{g'_{z}(1,1,1)}$
We showed that with the implicit function theorem and we took the derivative of the function $f(x,y,z)$ with respect to $z$ being a function $z(x,y)$ and we get $f'_{x}(1,1,1) = -2$. cool, no problem yet.
But in the other section, we are required to do the same thing, but this time show that
$(*)$ defines a function $y = y(x,z)$ around point $(1,1,1)$ then again, find $f'_{x}(1,1,1)$
We do the same thing for $y$ instead of $z$, so get a different derivative of the same function in the same point. We get $f'_{x}(1,1,1) = -1$ because we now respsected $y$ to be a function and not $z$
So I'm tackled by this because we get a different derivative for the same function in the same point, also how come on one hand $z$ is a function of $x$ and $y$ but if we look at it another way $y$ is a function of $x$ and $z$ yet I assume they can't both be a function in the same time ?
Writing $f'_x(1,1,1)$ for this is horrible notation, so no wonder you're confused...
What they do is define a composite function, let's call it $F$, by $$ F(x,y) = f(x,y,z(x,y)) . $$ This function has partial derivatives $F'_x(x,y)$ and $F'_y(x,y)$ which can be found using the chain rule, and it's $F'_x(1,1)$ that you're computing.
Next they define another composite function $$ G(x,z) = f(x,y(x,z),z) $$ and ask you to compute $G'_x(1,1)$.