You're looking for extrema of [tex]f(x,y)=5x+5y[/tex] subject to [tex]9x^2-9xy+9y^2=4[/tex].
The Lagrangian is
[tex]L(x,y,\lambda)=5x+5y+\lambda(9x^2-9xy+9y^2-4)[/tex]
with critical points where
[tex]L_x=5+\lambda(18x-9y)=0\implies\dfrac5\lambda=9y-18x[/tex]
[tex]L_y=5+\lambda(18y-9x)=0\implies\dfrac5\lambda=9x-18y[/tex]
[tex]L_\lambda=9x^2-9xy+9y^2-4=0[/tex]
[tex]L_x=0[/tex] and [tex]L_y=0[/tex] together tell you that
[tex]9y-18x=9x-18y\implies27x=27y\implies x=y[/tex]
Substituting [tex]y=x[/tex] into [tex]L_\lambda=0[/tex] gives you
[tex]9x^2-9x^2+9x^2-4=0\implies x^2=\dfrac49\implies x=\pm\dfrac23[/tex]
So you get two critical points, [tex]\left(-\dfrac23,-\dfrac23\right)[/tex] and [tex]\left(\dfrac23,\dfrac23\right)[/tex].
[tex]f\left(-\dfrac23,-\dfrac23\right)=-\dfrac{20}3[/tex] (min)
[tex]f\left(\dfrac23,\dfrac23\right)=\dfrac{20}3[/tex] (max)
