Too bad that your exam is so soon. With more time I might learn to read Georgian.
Let me guess what the problem asks for. We draw a total of $10$ lines, some (perhaps $0$) horizontal, the rest vertical. The question (I am guessing) now says that these lines divide the square into "equal" (presumably congruent) rectangles which are not squares. What is the maximum possible number of such rectangles?
Using the notation of Ross Millikan, let us see what happens if we draw $m$ horizontal lines and $n$ vertical lines, where $m+n=10$.
This procedure divides the square into $(m+1)(n+1)$ "rectangles".
We want to maximize $(m+1)(n+1)$, subject to the condition that $m+n=10$ and
$m \ne n$. (If $m=n$, we will have divided the square into squares.)
To put it in other terms, we want to maximize $xy$, subject to the following conditions:
$1$) $x$ and $y$ are unequal positive integers
There is general theory around this: $xy$ is maximized when $x$ and $y$ are as nearly equal as possible. But in an exam context, you can just quickly examine various possibilities in your mind. Best choice is $x=5$, $y=7$ (or the other way around) for a total of $35$.