So I got a bit stuck with the technicalities of this exercise.
Let $f(x,y,z)=\sqrt {x^2+y^2+z^2}$. Find $f$ minimal points under the constraints $x^2+y^2=1$ and $x^2-xy+y^2-z^2=1$ using Lagrange multipliers.
Honestly I had troubles extracting the critical points. I would greatly appreciate any assistance.
It may be helpful to start from a geometrical interpretation. The function $ \ f(x,y,z) \ $ is of course the distance of a point $ \ (x,y,z) \ $ from the origin. The surface $ \ x^2 \ + \ y^2 \ = \ 1 \ $ is a cylinder with its axis running along the $ \ z-$ axis and with "horizontal" circular cross-sections of radius 1. The other surface $ \ x^2 \ - \ xy \ + \ y^2 \ - \ z^2 \ = \ 1 \ $ is a hyperboloid of one sheet, with horizontal elliptical cross-sections having their major axes oriented 45º to the $ \ x- \ $ and $ \ y-$ axes and its axis also along the $ \ z-$ axis. The situation looks like this:
The pair of constraint equations taken "simultaneously" corresponds to the intersections of the cylinder and the hyperboloid, which are two closed curves, one of which is marked in yellow in the diagram above (the second one is on the "far side" of the surfaces). The problem is then to find the points on these curves which are closest to the origin, and to determine that minimal distance.
It will be more convenient in making the calculations for the Lagrange-multiplier method to use the "distance-squared" function $ \ \phi \ = \ x^2 \ + \ y^2 \ + \ z^2 \ $ ; since distance is non-negative, minimizing $ \ \phi \ $ will serve to minimize $ \ f \ $ . Depending upon whether you've learned the "basic" method for a single constraint function, or how the method is extended to multiple constraints, we can solve the problem in the following ways.
To use a single "multiplier", we must combine the two constraint equations to produce a single constraint function, so that we can set up the "Lagrange equations" implied by $ \ \nabla \phi \ = \ \lambda \ \nabla g \ $ . The gradient for the "distance-squared" function is $ \ \nabla \phi \ = \ \langle \ 2x \ , \ 2y \ , \ 2z \ \rangle \ $ . We can simply insert the equation for the cylinder into that for the hyperboloid to obtain
$$ (x^2 \ + \ y^2) \ - \ xy \ - \ z^2 \ = \ 1 \ \ \Rightarrow \ \ 1 \ - \ xy \ - \ z^2 \ = \ 1 \ \ \Rightarrow \ \ xy \ + \ z^2 \ = \ 0 \ \ . $$
From this, we can then assign the constraint function $ \ g(x,y,z) \ = \ xy \ + \ z^2 \ ( \ - \ 0 \ ) \ $ , the gradient of which is $ \ \nabla g \ = \ \langle \ y \ , \ x \ , \ 2z \ \rangle \ $ . The Lagrange equations are then
$$ \ 2x \ = \ \lambda \cdot y \ \ , \ \ 2y \ = \ \lambda \cdot x \ \ , \ \ 2z \ = \ \lambda \cdot 2z \ \ . $$
The last of these can be written as $ \ 2z \ ( \ \lambda \ - \ 1 \ ) \ = \ 0 \ $ , implying that either $ \ z \ = \ 0 \ $ or $ \ \lambda \ = \ 1 \ $ . In this first case, our second constraint equation tells us that $ \ xy \ + \ 0^2 \ = \ 0 \ \Rightarrow \ xy \ = \ 0 \ $ . Taken together with the first constraint, the equation of the cylinder, we find as critical points the four points on the "equator" of the hyperboloid
$$ (0, 1, 0) \ , \ (0, -1, 0) \ , \ (1, 0, 0) \ , \ (-1, 0, 0) \ \ , $$
all of which are at a distance 1 from the origin. In the second case, setting $ \ \lambda \ = \ 1 \ $ tells us nothing about $ \ z \ $ ; the other two Lagrange equations yield $ \ 2x \ = \ y \ , \ 2y \ = \ x \ \Rightarrow \ x \ = \ 0 \ \Rightarrow \ y \ = \ 0 \ $ . But this result does not fall on either surface. If we instead write
$$ \lambda \ = \ \frac{2x}{y} \ = \ \frac{2y}{x} \ \neq \ 1 \ , $$ neglecting the value of $ \ z \ $ for the present, we obtain $ \ 2x^2 \ = \ 2y^2 \ \Rightarrow \ x \ = \ \pm y \ $ . Inserting this into the cylinder equation implies that $ \ 2x^2 \ = \ 1 \ \Rightarrow \ x \ = \ \pm \frac{1}{\sqrt{2}} \ . $ We have two cases to consider here for the hyperboloid equation:
x = y -- $ \ \ \ x^2 \ - \ x \cdot x \ + \ x^2 \ - \ z^2 \ = \ 1 \ \ \Rightarrow \ \ \left(\pm \frac{1}{\sqrt{2}}\right)^2 \ - \ z^2 \ = \ 1 \ \ \Rightarrow \ \ z^2 \ = \ -\frac{1}{2} \ \ , $
which tells us that there are no such points on the hyperboloid;
x = -y -- $ \ \ \ x^2 \ - \ x \cdot (-x) \ + \ x^2 \ - \ z^2 \ = \ 1 \ \ \Rightarrow \ \ 3 \ \left(\pm \frac{1}{\sqrt{2}}\right)^2 \ - \ z^2 \ = \ 1 \ \ \Rightarrow \ \ z^2 \ = \ +\frac{1}{2} \ \ , $
showing us a second set of four critical points,
$$ ( \ \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \ ) \ , \ ( \ \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \ ) \ , \ ( \ -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \ ) \ , \ ( \ -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \ ) \ \ , $$
all of which lie at a distance $ \ \frac{3}{2} \ $ from the origin. [We can determine from this, incidentally, that this corresponds to $ \ \lambda \ = \ -2 \ $ .]
If we peer "down the throat" of the hyperboloid, we see half of the points in the two sets on the intersections of the cylinder and the hyperboloid. Those marked in red are the ones $ \ \frac{3}{2} \ $ units from the origin, the others in yellow are the points at unit distance, lying closest to the origin.
We had to "cheat" a little to work this all out with the single constraint function; it works a bit more cleanly with two. We now write the cylindrical constraint function $ \ g(x,y,z) \ = \ x^2 \ + \ y^2 \ - \ 1 \ $ [with $ \ \nabla g \ = \ \langle \ 2x \ , \ 2y \ , \ 0 \ \rangle \ $ ] and the hyperboloidal constraint function $ \ h(x,y,z) \ = \ x^2 \ - \ xy \ + \ y^2 \ - \ z^2 \ - \ 1 \ $ [ $ \ \nabla h \ = \ \langle \ 2x - y \ , \ 2y - x \ , \ -2z \ \rangle \ $ ]. The two-constraint Lagrange equations are
$$ \nabla f \ = \ \lambda \ \nabla g \ + \ \mu \ \nabla h \ \ \rightarrow $$
$$ 2x \ = \ \lambda \cdot 2x \ + \ \mu \cdot (2x \ - \ y) \ \ , \ \ 2y \ = \ \lambda \cdot 2y \ + \ \mu \cdot (2y \ - \ x) \ \ , \ \ 2z \ = \ \mu \cdot (-2z) \ \ . $$
From the equation in $ \ z \ $ , we conclude that either $ \ z \ = \ 0 \ $ , which gives us the first set of ("equatorial") critical points we found above, or else $ \ \mu \ = \ -1 \ $ . For this case, the other two Lagrange equations become
$$ 2x \ = \ \lambda \cdot 2x \ - \ (2x \ - \ y) \ \ \Rightarrow \ \ 4x \ - \ y \ = \ \lambda \cdot 2x \ \ , $$
$$ 2y \ = \ \lambda \cdot 2y \ - \ (2y \ - \ x) \ \ \Rightarrow \ \ 4y \ - \ x \ = \ \lambda \cdot 2y $$
$$ \Rightarrow \ \ \lambda \ = \ \frac{4x \ - \ y}{2x} \ = \ \frac{4y \ - \ x}{2y} \ \ \Rightarrow \ \ 8xy \ - \ 2y^2 \ = \ 8xy \ - \ 2x^2 \ \ \Rightarrow \ \ x \ = \ \pm y \ \ , $$
producing the other set of critical points, which are not at the minimal distance from the origin. With the two-constraint calculation, we obtain everything without needing to make the questionable omission of a condition we did earlier.