Answer:
a. -1/18
Explanation:
Differentiating implicitly, you have ...
y^2 +2xyy' -2y' +12y^2y' = 0
Solving for y', we get ...
y'(2xy -2 +12y^2) = -y^2
y' = -y^2/(2xy -2 +12y^2)
To make use of this, we need to know the value of x at y=1. Filling in y=1 into the given equation, we have ...
x -2 +4 = 6
x = 4 . . . . . . . . subtract 2
So, at the point (x, y) = (4, 1), the slope is ...
y' = -1/(8 -2 +12)
y' = -1/18
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The attached graph shows that the line with slope -1/18 appears to be tangent to the curve at (4, 1).