As a real function:

#y = xsqrt(1-x^2)#

is defined when #1-x^2 >= 0#, that is for #x in [-1,1]#.

So the function is continuous on a compact domain and Weierstrass' theorem ensures it has an absolute maximum and an absolute minimum.

Evaluate first the value of #y# and the limits of the domain:

#y(-1) = y(1) = 0#

Evaluate now the first derivative:

#dy/dx = x (d/dx sqrt(1-x^2)) + (d/dx x) sqrt(1-x^2)#

#dy/dx = x (-x/sqrt(1-x^2)) + sqrt(1-x^2)#

#dy/dx = sqrt(1-x^2) -x^2/sqrt(1-x^2)#

#dy/dx = (1-x^2 -x^2)/sqrt(1-x^2)#

#dy/dx = (1-2x^2 )/sqrt(1-x^2)#

and find the critical points of #y(x)# inside its domain solving the equation:

#dy/dx =0#

#(1-2x^2 )/sqrt(1-x^2) = 0#

#1-2x^2 = 0#

#x=+-1/sqrt2#

Looking at the expression of #y'(x)# we can see that the denominator is always positive, while the numerator is a second degree polynomial with leading negative coefficient. #y'(x)# is therefore negative in the interval outside the roots and positive in the interval between the roots.

We can then conclude that:

#y(x)# is decreasing in #(-1,-1/sqrt2)#

#y(x)# is increasing in #(-1/sqrt2,1/sqrt2)#

#y(x)# is decreasing in #(1/sqrt2,1)#

so that #x=-1/sqrt2# is a local minimum and #x=1/sqrt2# is a local maximum. Evaluating:

#y(-1/sqrt2) = (-1/sqrt2)sqrt(1-1/2) = -1/2 #

#y(1/sqrt2) = (1/sqrt2)sqrt(1-1/2) = 1/2 #

we can see that the minimum is lower and the maximum is higher than the values of #y(x)# at the boundaries, so these are also the absolute minimum and maximum.