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.

You have the equation $1=\frac{-2x}{4y}$ which is equivalent to $x=-2y$. Now, the points you are looking for should be on the ellipse $x^2+2y^2=1$ so you have the additional constraint:
$$(-2y)^2+2y^2=1 \;\;\;\Leftrightarrow\;\;\;y^2=1/6.$$
Therefore, $y=\pm\frac{\sqrt{6}}{6}$. In turn, $x=-2y=\mp\frac{\sqrt{6}}{3}$, so the two points you want are
$$
(\sqrt{6}/3,-\sqrt{6}/6)\;\;\;\mbox{and}\;\;\;(-\sqrt{6}/3,\sqrt{6}/6)
$$
