First, transform the variables into Laplace domain for dealing with algebraic rather than differential equations, which greatly simplifies the labor. And then properly re-route those two feedback branches to simplify the block diagram yet still having the same overall transfer function.
All three sub-blocks of the system are integrators whose transfer function is $H(s) = 1/s$. And your feedback paths have unity gain.
Your initial system diagram looks like this:
The transfer function is defined like: $$ H(s) = \frac{Y(s)}{U(s)} $$
In the first step, lets move the upper feedback path, which is added to the output of the first integrator, to the left adder node.
Note that the feedback signal was $Y(s)$ . When you move this feedback signal to the front of the first integrator block, you should modify the feedback signal properly so that the two configurations produce exactly the same signals. Therefore, label the new feedback signal as $Y(s) s$, (more generally $Y(s)/H_1(s)$, where $H_1(s)=1/s$). This results in the following block diagram:
Observe that the middle adder is left free, and can be removed:
Then consider the lower feedback path which samples its signal from the input of the last integrator whose output is $Y(s)$. Therefore, the lower feedback path signal is $sY(s)$. Hence we can label the lower feedback path with the signal $sY(s)$.
This way, we can (apparently!) remove the feedback lines, as we have found their resulting (effective) input signals, and obtain the following final (simplified) block diagram in which you can merge the three cascaded blocks.
Finally, this is a simple input-output relationship (with feedback). : $$ Y(s) = \frac{1}{s^3} \{ 2sY(s) + U(s)\} $$
from which you find the transfer function as :
$$Y(s) \{ 1 - \frac{2}{s^2} \} = \frac{U(s)}{s^3} $$
$$H(s) = \frac{Y(s)}{U(s)} = \frac{1}{s^3 - 2s} $$
