Derivative Problem. Checking My Answer

3xy^2+lny=6
Find the tangent line at (2,1)
y=-3/13x+9/13

3xy² + ln(y) = 6

Differentiate Implicitly:
3y² + 6xyy' + y'/y = 0
y'[6xy + 1/y] = -3y²
y' = -3y² / [6xy + 1/y]
y' = -3y³ / [6xy² + 1]
y'(2,1) = -3/[12 + 1] = -3/13

Tangent Line:
y - 1 = (-3/13)(x - 2)
y = (-3/13)x + 19/13

Yes, that is correct.