Well, both Dominic's and Ittay's answers are good but to add to Ittay's answer, Ittay is absolutely right that for any polynomial the Taylor expansion is the polynomial itself BUT in this case writing $f(x,y,z)$ as $xy^2z^3$ is the Taylor expansion of $f(x)$ around the origin. The OP I think wants to expand $f(x)$ around the point (1,0,-1) in which case the Taylor expansion isn't $xy^2z^3$.
Just to give an example, if you want to expand $g(x)=x^2$ around $x=0$ then the Taylor expansion is $g(x)=x^2$. But if you want the Taylor expansion around $x=1$, it would instead be $g(x)=(x-1)^2+2(x-1)+1$. And this case we can use the trick
The polynomials are identical. If you simplify the series you do get $x^2$ but the FORM is different because a Taylor polynomial expanded at $x=a$ must be written as a polynomial in $(x-a)$.
So for your original question $f(x,y,z)=xy^2z^3$ is NOT the Taylor expansion around (1,0,-1). Either use the definition of the multivariate Taylor expansion carefully...which as you correctly noted is quite hairy...or we can use the same trick again. Write
and then just expand this polynomial keeping the parenthesis grouped together and you will have a polynomial in $(x-1)$, $y$ which is the same as $(y-0)$, and $(z+1)$.