Let the equation of the line be $y = mx+c$. Since the line passes through $(1,1)$, we have $1 = m + c$ i.e. $c = 1 - m$. Hence, the equation of the line is $y = mx + 1 - m$.
Now we want the above line to be a tangent to $2x^2 + y^2 = 1$. This means the line should intersect the ellipse at exactly one point. Plug in $y = mx + 1 - m$ and the condition that the line intersects the ellipse at only one point means that the quadratic in $x$ must have only one root which means that the discriminant of the quadratic must be zero.
The quadratic we get is $2x^2 + (mx + 1 - m)^2 = 1$. Rearranging we get $$(m^2+2)x^2 + 2m(1-m)x + m^2 - 2m = 0$$
The discriminant is $4m^2(1-m)^2 - 4(m^2-2m)(m^2+2) = -4m(m-4)$
Setting this to zero, we get the two tangent from $(1,1)$ to the ellipse $2x^2 + y^2 = 1$ are
$$y= 1$$ $$y = 4x - 3$$