Question

Consider f and c below. f(x, y, z) = (y2z + 2xz2)i + 2xyzj + (xy2 + 2x2z)k,

Answer 1

$\dfrac{\partial f}{\partial x}=y^2z+2xz^2$
$f(x,y,z)=xy^2z+x^2z^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=2xyz+\dfrac{\partial g}{\partial y}=2xyz$
$\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$\dfrac{\partial f}{\partial z}=xy^2+2x^2z+\dfrac{\mathrm dh}{\mathrm dz}=xy^2+2x^2z$
$\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=0$

$\implies f(x,y,z)=xy^2z+x^2z^2+C$