The slope is #3#.
You could solve for #y#, then take the derivative and find the slope.
(I'll do that after the other method.)
I'm going to guess that this has come up in your study of implicit differentiation, so I'll do it that way.
We need to find #(dy)/(dx)# and evaluate at the given point. Rather than solve for #y# first, we'll leave the function implicit (hidden, implied) and differentiate using the chain rule.
In our thoughts, #y# is some function(s) of #x#, we haven't bothered to find the function,
but #x^2y^2# is really #x^2*("some function of x")^2#
Differentiating such an expression requires the product rule, the power rule and the chain rule. Note that:
the derivative of #("some function of x")^2# is #2*("some function of x")*("the derivative of the function")#. That is tedious, but we have a name for the function. It's called #y#. #d/(dx)(y^2)=2y(dy)/(dx)#.
Returning to the question:
#x^2y^2=9#, so
#d/(dx)(x^2y^2)=d/(dx)(9)#
#2xy^2+x^2 2y (dy)/(dx) = 0#, and
#xy^2+x^2 y (dy)/(dx)=0#.
#(dy)/(dx)= -color(red)(y)/x# (For this curve #x,y!=0#)
At #(-1, 3)#, we see that #(dy)/(dx)=3#
The slope of the tangent to the curve at the indicated point is #3#
Note
In the equation: #(dy)/(dx)= -color(red)(y)/x#, unlike earlier equations for #(dy)/(dx)#, we see not only #x#, but also #y#. That is the price we paid for not solving for #y# before differentiating.
#x^2y^2=9# implies #y^2=9/(x^2)#, so #y=+-3/x#, There are 2 branches to the graph. The branch that includes #(-1, 3)# has #y=-3/x#.
On this branch #(dy)/(dx)=3/x^2# (Note the absence of #y# in this equation.)
So, at the point of interest, #(dy)/(dx)=3/(-1)^2=3#.
