The height of a point $P$ of the ellipse is given by the intersection of the two circles of equations
$$\begin{cases}
(x+d/2)^2 + y^2 = d_1^2\\
(x-d/2)^2 + y^2 = d_2^2
\end{cases}$$
where $(-d/2,0), (d/2,0)$ are the points where the strings are attached. And $d_1+d_2=a$ is a constant (the length of the string) larger than $d$. You're looking to the maximum of $y$ and want to prove that it is obtained when $x=0$.
So
$$f(x,y)=\sqrt{(x+d/2)^2 + y^2} + \sqrt{(x-d/2)^2 + y^2}=a$$
$y$ is implicitely defined through $f$ as a function of $x$. You have
$$\begin{cases}
\frac{\partial f}{\partial x} = \frac{x+d/2}{\sqrt{(x+d/2)^2 + y^2}} + \frac{x-d/2}{\sqrt{(x-d/2)^2 + y^2}} \\
\frac{\partial f}{\partial y} = \frac{y}{\sqrt{(x+d/2)^2 + y^2}} + \frac{y}{\sqrt{(x-d/2)^2 + y^2}}
\end{cases}$$
The General formula for derivative of implicit function tells you that
$$y^\prime(x) = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$
Therefore $y^\prime(x)$ vanishes when $\frac{\partial f}{\partial x}$ vanishes, that is for $x=0$ as desired. And in that case, $d_1^2=d_2^2=d^2/4+y^2 = a^2/4$.