possible
The given point is on the curve. Find the lines that are (a) tangent and (b) normal to the curve at the given point.
$$6x^2+7xy+2y^2+13y-6=0,(-1,0)$$

Step 1 ：We are given the curve $6x^2+7xy+2y^2+13y-6=0$ and the point $(-1,0)$. We are asked to find the lines that are (a) tangent and (b) normal to the curve at the given point.

Step 2 ：To find the tangent and normal lines to the curve at a given point, we first need to find the derivative of the curve at that point. The derivative will give us the slope of the tangent line. The slope of the normal line is the negative reciprocal of the slope of the tangent line.

Step 3 ：We can find the derivative using implicit differentiation. The derivative of the given function is $12x+7y+y^{\prime }(7x+4y+13)$.

Step 4 ：Substituting the given point $(-1,0)$ into the derivative, we find that the slope of the tangent line is 2.

Step 5 ：The slope of the normal line is the negative reciprocal of the slope of the tangent line, which is -1/2.

Step 6 ：Now that we have the slopes of the tangent and normal lines, we can use the point-slope form of a line to find the equations of these lines. The point-slope form of a line is given by $y-y1=m(x-x1)$, where m is the slope and $(x1,y1)$ is a point on the line.

Step 7 ：Substituting the given point $(-1,0)$ and the slope of the tangent line into the point-slope form, we find that the equation of the tangent line is $y=2x+2$.

Step 8 ：Substituting the given point $(-1,0)$ and the slope of the normal line into the point-slope form, we find that the equation of the normal line is $y=-\frac{x}{2}-\frac{1}{2}$.

Step 9 ：Final Answer: The equation of the tangent line to the curve at the point $(-1,0)$ is $\boxed{{\displaystyle y=2x+2}}$ and the equation of the normal line to the curve at the point $(-1,0)$ is $\boxed{{\displaystyle y=-\frac{x}{2}-\frac{1}{2}}}$.