The steady states of u are, of course, the values of u that make du/dt equal to 0. Since \(\displaystyle \frac{du}{dt}= u(1- u)(1+ u)- Eu= u[(1- u)(1+ u)- E]\) one such steady state is u= 0. If u is not 0 we can divide by it to get \(\displaystyle (1- u)(1+ u)- E= 1- u^2- E= 0\) or \(\displaystyle u^2= 1- E\). As long as E is less than 1 the other two steady states are \(\displaystyle u= \sqrt{1- E}\) and \(\displaystyle u= -\sqrt{1- E}\). (If E= 1 or larger those three collapse to the single u= 0.) You have cut off the first part of the question so it is not clear what u means, biologically, so not clear whether u negative is "biologically relevant". If u is a "population" then only \(\displaystyle u= \sqrt{1- E}\) or u= 0 are "biologically relevant".
u* is defined as one of those steady states. Obviously, if u*= 0, Eu*(E)= 0 and there is no "yield" so only \(\displaystyle u*= \sqrt{1- E}\) is relevant. With \(\displaystyle u*= \sqrt{1- E}\) the yield is \(\displaystyle Eu*(E)= E\sqrt{1- E}= E(1- E)^{1/2}\). By the "product rule" the derivative with respect to E is \(\displaystyle (1- E)^{1/2}- \frac{1}{2}E(1- E)^{-1/2}\).
Setting that equal to 0 and multiplying both sides by \(\displaystyle (1- E)^{1/2}\), \(\displaystyle 1- E- \frac{1}{2}E=1-\frac{3}{2}E= 0\) so \(\displaystyle E= \frac{2}{3}\). The steady state population is \(\displaystyle u= \sqrt{1- E}= \sqrt{1- \frac{2}{3}}= \sqrt{\frac{1}{3}}= \frac{\sqrt{3}}{3}\).
