We can think of the two surfaces as "equi-potential" surfaces of f(x,y,z)= z- x^2- y^2 and g(x,y,z)= x+ y+ z- e^{xyz}. Their gradients, \nabla f= -2x\vec{i}- 2y\vec{j}+ \vec{k} and \nabla g= (1- yze^{xyz})\vec{i}+ (1- xze^{xyz})\vec{j}+ (1- xye^{xyz})\vec{k} are perpendicular to the surfaces and perpendicular to the tangent planes at each point.
In particular, at (0, 0, 1), \nabla f(0, 0, 1)= \vec{k}] so the tangent plane is 0(x- 0)+ 0(y- 0)+ 1(z- 1)= 0 or z= 1. And \nabla g(0, 0, 1)= \vec{i}+ \vec{j}+ \vec{k} so the tangent plane is 1(x- 0)+ 1(y- 0)+ 1(z- 1)= x+ y+ z- 1= 0 or x+ y+ z= 1. The angle between those two planes is the same as the ange between the two perendicular vectors and you can use \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta) to find that angle.
The length of \vec{k} is 1, the length of \vec{i}+ \vec{j}+ \vec{k} is \sqrt{3}, and their dot product is 1. So 1(\sqrt{3})cos(\theta)= 1.