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I'm pretty sure that what you're looking for is: Either the slope or the full equation

of the line that is tangent to the curve represented by that messy equation, at

the point (1, 1).

If that's so, then . . .

-- You'd have to start out by knowing where the plus, minus, and 'equals' signs

are in the equation.

From there, what I would do is:

-- Take the implicit first derivative of the equation, and solve it for y' .

That gives you the slope of the curve as a function of points (x, y).

-- Use it to find the slope of the curve at the point (1, 1).

-- You know that the slope of the line that's tangent at that point is the

negative reciprocal of the slope of the curve. Now you have the slope of

the tangent line. If that's all you need, then you're done. Retreat and declare

victory.

-- If you need the equation of the tangent line, that's no problem. You have

its slope AND a point on it ==> (1, 1). Using that info to find the equation

of the line is supposed to be a piece o' cake for you by now.

=============================

Wait. Don't go away.

It looks to me as if the equation is

3x3 + 7x2y + 7xy2 + 2y3 = 19

just because the point (1, 1) is on it.

Do not trust anything I say past this point. I have already wasted too much time on this one and I am not going to check my work. But I did differentiate that mess, and I got the following:

Slope of the equation at (1, 1) . . . -10/9

Slope of the line tangent to the graph at (1, 1) . . . +0.9

Equation of the line . . . Y = 0.9x + 0.1

Take it for what it's worth, and Good Luck.

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Q: How do you find the tangent of the line at the curve at the point 11 using the equation 3x3 7x2y 7xy2 2y319?
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