Find the equation of the tangent line to the curve at the given point. \[ 13 x-x y+y^{2}=13 \text { at }(1,1) \]

To find the equation of the tangent line to the curve given by \( 13x - xy + y^2 = 13 \) at the point \((1,1)\), we'll first use implicit differentiation. Differentiating both sides of the equation with respect to \( x \): \[ \frac{d}{dx}(13x) - \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = 0 \] This leads to: \[ 13 - (y + x\frac{dy}{dx}) + 2y\frac{dy}{dx} = 0 \] Rearranging gives: \[ x\frac{dy}{dx} - 2y\frac{dy}{dx} = 13 - y \] Factoring out \(\frac{dy}{dx}\): \[ \frac{dy}{dx}(x - 2y) = 13 - y \] Now, solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{13 - y}{x - 2y} \] Now substitute \( (1,1) \): \[ \frac{dy}{dx} = \frac{13 - 1}{1 - 2(1)} = \frac{12}{1 - 2} = \frac{12}{-1} = -12 \] The slope of the tangent line at \((1,1)\) is \(-12\). Now, using the point-slope form \( y - y_1 = m(x - x_1) \): \[ y - 1 = -12(x - 1) \] Simplifying gives the equation of the tangent line: \[ y = -12x + 12 + 1 \implies y = -12x + 13 \] So, the equation of the tangent line at the point \((1,1)\) is: \[ y = -12x + 13 \]