Let S = x² + y²

To reduce the number of unknowns from 3 to 2, we solve 6x+2y=38 for y:
                                                          2y=38-6x
                                                           y=19-3x
Substituting 19-3x for y in S = x² + y²

S  = x² + (19-3x)²

Take the derivative:

S' = 2x + 2(19-3x)(-3)
S' = 2x - 6(19-3x)

Set the derivative equal to 0:

    2x - 6(19 - 3x) = 0
    2x -  114 + 18x = 0
          20x - 114 = 0
                20x = 114
                  x = 114/10
                  x = 57/10

Check to see if it's a minimum.  We take the second derivative

S" = 2 - 6(-3)
S" = 2 + 18
s" = 20

S" is not 0, so the second derivative test succeeds. It's positive 
which means that it's concave upward there, which means that it is
a minimum.

Now we find the y-value by substituting in y=19-3x

y=19-3x
y=19-3(57/10)
y=190/10-171/10
y=19/10

Answer: (57/10, 19/10)

Edwin

It can be solved by Algebra, without Calculus.


From 6x+2y = 38 express  y = 19-3x  and substitute it into 


    x^2 + y^2 = x^2 + (19-3x)^2 = x^2 + 19^2 - 2*19*3x + 9x^2 = 10x^2 - 2*57x + 19^2.   (1)


Now, the quadratic function (1)  has a minimum at


    x = "  " =  = .


That's all.