partial differentiation lagrange multiplier stuck at finding the solutions of equations

Question

$f(x,y) = \frac{x}{1+y^2}$,S=${(x,y);\frac{x^2}{4}+\frac{y^2}{9} \le 1}$
find maximum and mininmum of the function over the closed and bounded set S.
okay here i use lagrange multipliers, but i always confuse when i try to find the solution for the equations. the lagrange equation after partial differentiation:
1. $\frac{1}{1+y^2}=\frac{x\lambda}{2}$
2. $\frac{-2xy}{(1+y^2)^2}=\frac{2y\lambda}{9}$
3. $9x^2+4y^2=36$
then the solution said. one of the solutions is $y=0$ but if i put 1 to 2 and substitute and eliminate lambda, i got $x=\sqrt \frac{-40}{9}$ and $y=\sqrt 19$ and how to produce $\frac{-\lambda x^3y}{2}=\frac{2\lambda y}{9}$ ??
i was so confuse when trying to find the solutions of equations, and here is it wrong to eliminate $\lambda$?

Answer 1

Considering $$f=\frac{x}{y^2+1}-\lambda \left(\frac{x^2}{4}+\frac{y^2}{9}-1\right)$$ as you wrote $$f'_x=\frac{1}{y^2+1}-\frac{\lambda x}{2}=0\tag 1$$ $$f'_y=-\frac{2 x y}{\left(y^2+1\right)^2}-\frac{2 \lambda y}{9}=0\tag 2$$ $$f'_\lambda =\frac{x^2}{4}+\frac{y^2}{9}-1=0\tag 3$$ From $(1)$, eliminate $x$ to get $$x=\frac{2}{\lambda \left(y^2+1\right)}$$ Replace in $(2)$ and simplify to get $$f'_y=-\frac{2 y \left(\lambda ^2 \left(y^2+1\right)^3+18\right)}{9 \lambda \left(y^2+1\right)^3}$$ making $y=0$ an obvious solution. Now, you are then left with $(3)$.

Answer 2

Per your question, first square both sides of the top equation, and then multiply both sides of this new equation by $-2xy$ and you have this new equation with the second old equation have the same left side, then you equate their right sides which is the equation you wrote above.