The given point is on the curve. Find the lines that are a. tangent and $\mathbf{b}$. normal to the curve at the given point.
\[
x^{2}+x y-y^{2}=-9,(3,6)
\]
a. Give the equation of the tangent line to the curve at the given point
\[
y=
\]
b. Give the equation of the normal line to the curve at the given point.
\[
y=
\]

Step 1 ：Given the function \(x^{2}+x y-y^{2}=-9\) and the point \((3,6)\), we need to find the tangent and normal lines to the curve at this point.

Step 2 ：First, we find the derivative of the function using implicit differentiation. The derivative of the function is \(2x + y\).

Step 3 ：Substitute the given point into the derivative to find the slope of the tangent line. The slope of the tangent line is 12.

Step 4 ：The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the slope of the tangent line. The slope of the normal line is \(-\frac{1}{12}\).

Step 5 ：Use the point-slope form of a line, \(y - y1 = m(x - x1)\), where m is the slope and \((x1, y1)\) is the given point, to find the equations of the tangent and normal lines.

Step 6 ：The equation of the tangent line to the curve at the given point is \(y = 12x - 30\).

Step 7 ：The equation of the normal line to the curve at the given point is \(y = -\frac{1}{12}x + \frac{25}{4}\).

Step 8 ：Final Answer: The equation of the tangent line to the curve at the given point is \(\boxed{y = 12x - 30}\).

Step 9 ：Final Answer: The equation of the normal line to the curve at the given point is \(\boxed{y = -\frac{1}{12}x + \frac{25}{4}}\).