## University Calculus: Early Transcendentals (3rd Edition)

$xe^{y+2z}+C$
Here, $f=xe^{y+2z}+g(y,z)$ and $g(y,z)=0$ or, $g(y,z)=h(z)=0$ Then $f=xe^{y+2z}+h(z)$ This implies that $f_z=2xe^{y+2z}+h'(z)=2xe^{y+2z}$ Hence, $f=xe^{y+2z}+C$