Looking at this question again, I don't understand why you're told "for y=11". That doesn't seem relevant at all... So you can disregard the answer I posted a few minutes ago on your other question.
a) With [tex]x^2=-2+y+5\cos y[/tex], differentiate both sides with respect to [tex]x[/tex] to get
[tex]2x=\dfrac{\mathrm dy}{\mathrm dx}-5\sin y\dfrac{\mathrm dy}{\mathrm dx}[/tex]
[tex]\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x}{1-5\sin y}[/tex]
b) The point P occurs at [tex]x=2[/tex], which corresponds to a [tex]y[/tex]-coordinate of
[tex]4=-2+y+5\cos y\implies y\approx4.928[/tex]
The slope of the line tangent to this point is approximately
[tex]\dfrac{\mathrm dy}{\mathrm dx}\approx\dfrac{2(2)}{1-5\sin(4.928)}\approx0.68[/tex]
so the equation of the tangent line is approximately
[tex]y-4.928=0.68(x-2)\implies y=0.68x+3.57[/tex]
c) The tangent line to the graphed curve is vertical when [tex]\dfrac{\mathrm dy}{\mathrm dx}[/tex] is undefined. This happens when [tex]1-5\sin y=0[/tex], or [tex]y=\pi-\sin^{-1}\dfrac15+2n\pi[/tex] and [tex]y=\sin^{-1}\dfrac15+2n\pi[/tex] where [tex]n[/tex] is any integer.
In case you're not sure where the general solution came from: We have
[tex]\sin y=\dfrac15[/tex]
which has an infinite number of solutions. [tex]\sin^{-1}\dfrac15[/tex] is one of them, which we obtain by taking the inverse sine of both sides of this equation. Since [tex]\sin(\pi-x)=\sin x[/tex], we also know that [tex]\pi-\sin^{-1}\dfrac15[/tex] is a solution. And since [tex]\sin(x+2n\pi)=\sin x[/tex] for integers [tex]n[/tex], we also know that we can add any multiple of [tex]2\pi[/tex] to these two solutions to get infinitely more solutions.
