Use the method of Lagrange multipliers to find the ends with restrictions of the function $f(x,y)=xy$, subject to $x^2+y^2=2$
$\textit{Sol:}$
Form the equations system
$$\left\lbrace
\begin{array}{ll}
y=\alpha(2x) \\
x=\alpha(2y) \\
x^2+y^2-2=0
\end{array}
\right.$$
Replacing row 1 and 2 in row 3 $$\begin{align}(\alpha 2x)^2+(\alpha 2y)^2-2 &= 0 \\ \alpha &= \pm \sqrt{\frac{1}{2x^2+2y^2}} \end{align}$$ But $\alpha$ is in terms of $x$ and $y$. What should I do to reach a solution?
There are basically two kinds of "routine" Lagrange multipliers problems. In one, you start by solving for the Lagrange multiplier (as a number), and then plug that into the Lagrange conditions to find the solution. In the other, you cancel out the Lagrange multiplier from the problem altogether and then solve for the solution from there.
This problem is easier to do the second way. That's because you can divide the first equation by the second to get
$$\frac{y}{x}=\frac{x}{y}$$
hence $y^2=x^2$. Plugging that into the constraint gives the solution.
That said, following your approach, you can substitute in $x^2+y^2=2$ from the constraint to get the value of $\alpha$ (which amounts to approaching the problem the first way I mentioned).