Let F(x,y,z)=xz²-yz+cos(xy)-1.
∂F/∂x=z²-ysin(xy), ∂F/∂y=-z-xsin(xy), ∂F/∂z=2xz-y.
The tangent plane is:
∂F/∂x(x-a)+∂F/∂y(y-b)+∂F/∂z(y-c)=0, where (a,b,c)=(0,0,1) and the partial derivatives are at (0,0,1), so:
∂F/∂x=1, ∂F/∂y=-1, ∂F/∂z=0, making the tangent plane:
x-y=0.
When t=1, r=(0,0,1) which is the tangent point.