Answer
Maximum: $f(2,1,0)=5$ and Minimum: $f(0,-1,0)=1$
Work Step by Step
Use Lagrange Multipliers Method:
$\nabla f(x,y,z)=\lambda_1 \nabla g(x,y,z)+\lambda_2 \nabla h(x,y,z)$
This yields $\nabla f=\lt 2x,2y,2z \gt$ and $\lambda_1 \nabla g(x,y,z)+\lambda_2 \nabla h(x,y,z)= \lt \lambda_1,-\lambda_1+2\lambda_2y,-2\lambda_2 z \gt$
Using the constraint condition $x-y=1$we get, $x=2/3 ,y=-1/3$
If we use the above value of $x,y$ in second constraint $y^2-z^2=1$, we will get an invalid value.
Thus, we have to consider $z=0$; after solving, we get $x=2,0$ and $y=1,-1$
Hence, Maximum: $f(2,1,0)=5$ and Minimum: $f(0,-1,0)=1$
