x²z + y²z + 4xy = 40,

x² + y² + xyz = 20

If we devide both the equations by xy we get

z*(x/y+y/x) +4 = 40/xy and

(x/y+y/x) +z = 20/xy

(40/xy-4)*z = (20/xy-z)*2

Simple solution is z= 2 but then x,y = 2*5^0.5

If we add & substract 2xyz in equation 1 & 2xy in equation 2 we get

z* ( x+y)^2 +xy*(4-2z)=40

(x+y)^2 + xy( z-2)=20

from which we get

xy*( 4-2z-z^2+2z)=40-20z

xy*(2-z)(2+z)=20(2-z)

xy(2+z)=20

if xy is 1 then z=18

if xy is 2 then z=8 All interger solution is not possible

if xy is 4 then z=3

if xy is 5 then z=2 This also does not have all integer solution

if xy is 10 z=0 This also does not have all integer solution