## Calculus 8th Edition

The solution is $$f_z(e,1,0)=1.$$
Lets first find the derivative: $$f_z=\Big(x^{yz}\Big)'_z=\Big(\big(x^y\big)^z\Big)'_z=(x^y)^z\ln x^y=x^{yz}y\ln x,$$ where we used the fact that $x^{yz}=(x^y)^z$ and also the logarithmic rule $\ln x^y=y\ln x$. Now let us calculate by direct substitution $x=e$, $y=1$ and $z=0$: $$f_z(e,1,0)=e^{1\cdot0}=e^0=1.$$