Answer
$$ x \space e^{y+2z}+c $$
Work Step by Step
We have $$ f=x \space e^{y+2z}+g(y,z) \\g(y,z)=0 \implies g(y,z)=h(z)=0$$
This yiedls: $$ f=xe^{y+2z}+h(z)$$
Now, $$ f_z=2xe^{y+2z}+h'(z)=2xe^{y+2z}$$
So, $$ f=x \space e^{y+2z}+c $$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 16: Integrals and Vector Fields - Section 16.3 - Path Independence, Conservative Fields, and Potential Functions - Exercises 16.3 - Page 966: 9
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