On The Construction Of An Ellipse

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, $d_1^2=d_2^2=d^2/4+y^2 = a^2/4$.

One approach which is somewhat roundabout but avoids calculus would be:

First prove that an ellipse is a squished circle. This shows that the ellipse has a single unique highest point (because certainly a circle does, and squishing the $y$-coordinates uniformly everywhere doesn't change which points are higher than others).

However, if the point with distances $d_1$ and $d_2$ is highest, then by symmetry the point with distances $d_2$ and $d_1$ is just as high. And if $d_1\ne d_2$ they are different points and can't both be highest, a contradiction.