Question 638050
In order to solve this, we first have to expand.  In other words, since everything in the parenthesis is being squared, we would have to multiply what's in the parenthesis, by itself:

{{{(3x^(1/2)-y^(1/2))^2}}}={{{(3x^(1/2)-y^(1/2))*(3x^(1/2)-y^(1/2))}}}

To expand, we will use the FOIL method(firsts x firsts)+(outsides x outsides)+(insides x insides)+(lasts x lasts):

{{{(3x^(1/2)*3x^(1/2))+(3x^(1/2)*1y^(1/2))-(1y^(1/2)*3x^(1/2))-(1y^(1/2)*-1y^(1/2)))}}}

For this problem, we will be using two laws of exponents: 

1. {{{x^m*x^n=x^(m+n)}}}
2. {{{x^(m/n)=root(n,m)}}}

So, as we are multiplying the "firsts", we will use the 1st law:

{{{3x^(1/2)*3x^(1/2)=9*x^((1/2)+(1/2))=9*x}}}

Next, we will multiply our "outsides", using both laws:

{{{-3x^(1/2)*1y^(1/2)=-3x^(1/2)*1y^(1/2)=-3*sqrt(x)*sqrt(y)}}}

Then our "insides", again using both laws:

{{{-y^(1/2)*3x^(1/2)=-1sqrt(y)*3sqrt(x)}}}

And add them together using both of the above laws of exponents:

{{{-3*sqrt(x)*sqrt(y)-1sqrt(y)*3sqrt(x)=-6sqrt(x*y)}}}

Finally, we will multiply the "lasts", using the first law:

{{{-1y^(1/2)*-1y^(1/2)=y^((1/2)+(1/2))=y}}}

Now we will add everything together:

{{{9*x-6sqrt(x*y)+y}}}, which is our final answer