Find the points on the ellipse $x^2+2y^2=1$ where the tangent line has slope $1$. I did the following:
$$2x+4y\cdot y'=0\\y'=\frac{-2x}{4y}$$
Now I guess I should take all pairs $(x,y)$ such that:
$$\frac{-2x}{4y}=1$$
And I found: $\left(1,-\cfrac{1}{2}\right),\left(-1,\cfrac{1}{2}\right)$
And then I tried to do:
$$x^2+2y^2=1\\ y=\frac{\sqrt{1-x^2}}{\sqrt{2}}$$
And then derive $\frac{\sqrt{1-x^2}}{\sqrt{2}}$, this yielded the pairs: $\{(\sqrt{\frac{2}{3}},-\frac{1}{\sqrt{6}}),(-\sqrt{\frac{2}{3}},\frac{1}{\sqrt{6}})\}$
I don't see what I did wrong in the implicit differentiation method.

we have $x=-2y$ and we get $$4y^2+2y^2=1$$ thus we get $$y=\pm\sqrt{\frac{1}{6}}$$
