#y= 2xsqrt(9-x^2)#

First let's find the domain of #y#

#y in RR# where #(9-x^2)>= 0 -> absx<= 3#

Hence the domain of #y# is: #-3<=x<=+3#

Next we will find the turning points of #y# using #y'=0#

#y = 2x(9-x^2)^(1/2)#

Applying the product rule and chain rule

#y'# = #2x*1/2*(9-x^2)^(-1/2) * (-2x) + 2*(9-x^2)^(1/2)#

#= 2(9-x^2)^(1/2) - (2x^2)/((9-x^2)^(1/2))#

#= (2(9-x^2)-2x^2)/(sqrt(9-x^2))#

#= (18-4x^2)/sqrt(9-x^2)#

For critical #y#:

# (18-4x^2)/sqrt(9-x^2)=0 -> 18-4x^2=0#

#2x^2 =9 -> x= +- sqrt(9/2)#

#x= +-3/sqrt2 = +-(3sqrt2)/2#

This question asks for the intervals of #x# of increasing and decreasing #y# using the #y'#. To avoid using the second derivative, it is now helpful to observe the graph of #y# below:

graph{2xsqrt(9-x^2) [-20.28, 20.26, -10.13, 10.15]}

Since we know the turning points are where #x= +-(3sqrt2)/2# and that the domain of #y# is #x: [-3, +3]#

We can see that #y# is increasing for #x: (-(3sqrt2)/2, +(3sqrt2)/2)#

and #y# is decresing for #x: [-3, -(3sqrt2)/2) uu (+(3sqrt2)/2, +3]#