Maximize $\boxed{\mathbf{x}+\mathbf{y}}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$ Maximize $\mathbf{x}+\mathbf{y}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$

My approach
$\frac{x^{2}}{1 / 2}+\frac{y^{2}}{1 / 3} \leq 1$
Let $z=x+y$
$\mathrm{Now}, 4 \mathrm{x}+6 \mathrm{y} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{2 x}{3 y}$
$2 x^{2}+3 y^{2}=1$
What to do next? Any suggestion or Hint would be greatly appreciated!

Alternatively, the objective function $z=x+y$ will achieve its maximum when the contour line $y=-x+z$ will touch the ellipse $2x^2+3y^2=1$ from above. So, the slope of the tangent line must be equated to the slope of the contour line:
$$4x_0^2+6y_0^2y'=0\Rightarrow y'=-\frac{2x_0}{3y_0}=-1\Rightarrow x_0=\frac32y_0$$
Now plug this into the equation of ellipse:
$$2\cdot \left(\frac32y_0\right)^2+3y_0^2=1\Rightarrow y_0=\sqrt{\frac2{15}}\Rightarrow x_0=\sqrt{\frac3{10}}.$$
Hence, the maximum is $z(x_0,y_0)=\sqrt{\frac3{10}}+\sqrt{\frac2{15}}$.
