Find the equation of the tangent line to the curve (a lemniscate) $2\left(x^{2}+y^{2}\right)^{2}=25\left(x^{2}-y^{2}\right)$ at the point $(3,-1)$. The equation of this tangent line can be written in the form $y=m x+b$ where $m$ is: and where $b$ is:

Step 1 ：Given the equation of the curve (a lemniscate) \(2(x^{2}+y^{2})^{2}=25(x^{2}-y^{2})\) and the point \((3,-1)\), we are to find the equation of the tangent line at this point.

Step 2 ：The equation of a tangent line to a curve at a given point can be found using the derivative of the function. However, the given equation is not a function of x or y alone, but a relation between x and y. Therefore, we need to use implicit differentiation to find the derivative.

Step 3 ：Using implicit differentiation, we find the derivative of the given equation to be \(8x(x^{2} + y^{2}) - 50x - y(8y(x^{2} + y^{2}) + 50y)\).

Step 4 ：The derivative dy/dx at a point gives the slope of the tangent line at that point. Substituting the given point \((3,-1)\) into the derivative, we find the slope \(m\) to be \(-\sqrt{-\sqrt{2929}/8 - 49/8}\).

Step 5 ：Once we have the slope, we can use the point-slope form of a line to find the equation of the tangent line. Substituting the slope and the given point into the point-slope form, we find the y-intercept \(b\) to be \(-1 + 3\sqrt{-\sqrt{2929}/8 - 49/8}\).

Step 6 ：Final Answer: The slope of the tangent line, \(m\), is \(\boxed{-\sqrt{-\sqrt{2929}/8 - 49/8}}\) and the y-intercept, \(b\), is \(\boxed{-1 + 3\sqrt{-\sqrt{2929}/8 - 49/8}}\).