<s>[INST] Finding minimum value of $ \frac{x^2 +y^2}{y} $ 
Finding the minimum value of $\displaystyle \frac{x^2 +y^2}{y}.$ where $x,y$ are 
  real
   numbers satisfying $7x^2 + 3xy + 3y^2 = 1$

Try: Equation $7x^2+3xy+3y^2=1$ represent Ellipse
with center is at origin.
So substitute $x=r\cos \alpha $ and $y=r\sin \alpha$ 
in $7x^2+3xy+3y^2=1$
$$3r^2+4r^2\cos^2 \alpha+3r^2\sin \alpha \cos \alpha =1$$
$$3r^2+2r^2(1+\cos 2 \alpha)+\frac{3r^2}{2}\sin 2 \alpha =1$$
$$8r^2+r^2(4\cos 2 \alpha+3\sin \alpha)=2$$
So $$r^2=\frac{2}{8+(4\cos 2 \alpha+3\sin \alpha)}$$
$$\frac{2}{8+5}=\frac{2}{13}\leq r^2\leq \frac{2}{8-5}=\frac{2}{3}$$
we have to find minimum of $$\frac{x^2+y^2}{y}=\frac{r}{\sin \alpha}$$
How can i find it, could some help me 
 [/INST] Lagrange function is
$$L=k\, \left( 3 {{y}^{2}}+3 x y+7 {{x}^{2}}-1\right) +\frac{{{y}^{2}}+{{x}^{2}}}{y}$$
Solve system:
$$L'_x=0,\quad L'_y=0,\quad L'_k=0.$$
We get two solutions
$$k=1/4,y=-2/5,x=-1/5$$ $$k=-1/4,y=2/5,x=1/5$$
Answer:
$$f_{min}=f\left(-\frac{1}{5},-\frac{2}{5}\right)=-\frac{1}{2}$$
with CAS Maxima:

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