Find slope of the tangent line to the curve at a point?

I would use a derivative, and then point-slope form. I see the question as
(2+(x^2)*(y^2))^(1/2) - 3xy = -31.9

First, derive implicitly (# stands for dy/dx):
(2xyy+2yxx#)/(2*(2+xxyy)^(1/2)) - 3(x#+y) = 0
Isolate the # on one side by moving the 3(x#+y) to the other side. Ten multiply by the denominator in the first expression. Then move the 2yxx# to the other side, while taking care of any other variables. I think you can take it from there. Sorry this isn't a thorough explanation, and it's not the best I can do, but it should suffice. Have fun!

dy/dx = - { x / √ (2 + x²y² ) - 3y } / { y / √(2 + x² y² ) - 3x }....put in x = 2 & y = 8

Get a calculator.