$$J(x,y,z)=x^2+\frac{y^3}{3}+z$$ subject to $x-y+z=0$ and $2x+y=z$. Find the stationary points and determine the nature of each one.
Attempt:
By using $\nabla J=\lambda \nabla g+\mu \nabla h$ where $g,h$ are the two constaints:
$2x=\lambda+\mu$
$y^2=-\lambda+\mu$
$1=\lambda-\mu$
I don't know how to continue from here because I get $y^2=-1$ from these equations
You don't have stationary points. Let's look at the constraints. We can write them as $$x=0\\y=z$$ If you plug these into the original expression, you would need to minimize $$f(y)=J(0,y,y)=\frac{y^3}3+y$$ Taking the derivative with respect to $y$ and setting it to $0$, you get your expression$$y^2+1=0$$This has no real solutions. This function is monotonically increasing with $y$, and it does not have any stationary point.