If minus, x, y, plus, 1, plus, y, squared, equals, minus, 3, x, cubed−xy+1+y 2 =−3x 3 then find the equations of all tangent lines to the curve when x, equals, minus, 1, .x=−1.

The question seems to be a bit jumbled, but I'll try to interpret it as best as I can. It seems like you're asking for the equations of the tangent lines to a curve defined by a certain equation at the point where x = -1.

First, we need to clarify the equation of the curve. From what you've written, it seems like it might be -x + y + 1 + y^2 = -3x^3. However, it's a bit unclear because of the way the equation is written. Could you please confirm if this is the correct equation?

Assuming this is the correct equation, the next step would be to find the derivative of the equation with respect to x. This will give us the slope of the tangent line at any point on the curve.

The derivative of -x + y + 1 + y^2 with respect to x is -1 + dy/dx + 2y*(dy/dx) = -9x^2. Here, dy/dx represents the derivative of y with respect to x.

We can solve this equation for dy/dx to find the slope of the tangent line at any point on the curve.

Then, we substitute x = -1 into this equation to find the slope of the tangent line at the point where x = -1.

Finally, we use the point-slope form of the equation of a line, y - y1 = m(x - x1), to find the equation of the tangent line. Here, (x1, y1) is the point on the curve where x = -1, and m is the slope of the tangent line at this point.

Please note that this is a general approach and the actual calculations might be different depending on the correct form of the original equation.