By the product rule,
[tex]f(x)=x^m(1-x)^n[/tex]
has derivative
[tex]f'(x)=mx^{m-1}(1-x)^n-nx^m(1-x)^{n-1}=x^{m-1}(1-x)^{n-1}(m(1-x)-nx)[/tex]
[tex]\implies f'(x)=x^{m-1}(1-x)^{n-1}(m-(m+n)x)[/tex]
Critical points occur where the derivative vanishes:
[tex]x^{m-1}(1-x)^{n-1}(m-(m+n)x)=0[/tex]
[tex]x^{m-1}=0\text{ or }(1-x)^{n-1}=0\text{ or }m-(m+n)x=0[/tex]
[tex]\implies x=0\text{ or }x=1\text{ or }x=\dfrac m{m+n}[/tex]