Use implicit differentiation to find an equation of the tangent line to the graph of the equation at the given point. \[ x^{2}+x \arctan (y)=y-1,\left(-\frac{\pi}{4}, 1\right) \] \( y=\square \)

To find the equation of the tangent line using implicit differentiation, we start with the given equation: \[ x^{2} + x \arctan(y) = y - 1. \] Firstly, we differentiate both sides with respect to \( x \): \[ 2x + \left( \arctan(y) + x \cdot \frac{1}{1+y^2} \cdot \frac{dy}{dx}\right) = \frac{dy}{dx}. \] Now package all the \( \frac{dy}{dx} \) terms on one side: \[ 2x + \arctan(y) = \left(1 - \frac{x}{1+y^2}\right) \frac{dy}{dx}. \] Now, we substitute the point \( \left(-\frac{\pi}{4}, 1\right) \): \[ 2 \left(-\frac{\pi}{4}\right) + \arctan(1) = \left(1 - \frac{-\frac{\pi}{4}}{1+1}\right) \frac{dy}{dx}. \] Recognizing that \( \arctan(1) = \frac{\pi}{4} \) and simplifying: \[ -\frac{\pi}{2} + \frac{\pi}{4} = \left(1 + \frac{\pi}{8}\right) \frac{dy}{dx}, \] which leads to: \[ -\frac{\pi}{4} = \left(1 + \frac{\pi}{8}\right) \frac{dy}{dx}. \] Solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -\frac{\frac{\pi}{4}}{1 + \frac{\pi}{8}} = -\frac{\pi/4}{(8+\pi)/8} = -\frac{2\pi}{8+\pi}. \] Now, we have the slope \( m = -\frac{2\pi}{8+\pi} \). Finally, using the point-slope form of the line, the equation of the tangent line at the point \( \left(-\frac{\pi}{4}, 1\right) \) is: \[ y - 1 = -\frac{2\pi}{8+\pi} \left(x + \frac{\pi}{4}\right). \] Thus, the equation simplifies to: \[ y = -\frac{2\pi}{8+\pi}x + \left(1 + \frac{2\pi \cdot \frac{\pi}{4}}{8+\pi}\right). \] So the final answer for \( y \) is: \[ y = -\frac{2\pi}{8+\pi}x + \text{(a constant depending on } \pi \text{)}. \]