What is the equation of the tangent line to xy-3y^2=1 at the point (4 1)?

Differentiating implicitly, you have
.. y +x*dy/dx -6y*dy/dx = 0
.. dy/dx = y/(6y -x)
At the given point, this is
.. dy/dx = 1/(6*1 -4) = 1/2

The point-slope form of the equaton for the line is
.. y -1 = (1/2)(x -4)
or
.. x -2y = 2