Find the equation of the tangent line at the given point on the curve. \[ e^{2 x^{2}+2 y^{2}}=x e^{29 y}-y^{2} e^{\frac{58 x}{5}} ;(5,2) \]

Step 1 ：We are given the function \(e^{2 x^{2}+2 y^{2}}=x e^{29 y}-y^{2} e^{\frac{58 x}{5}}\) and we are asked to find the equation of the tangent line at the point (5,2).

Step 2 ：To find the equation of the tangent line at a given point on the curve, we need to find the derivative of the function at that point. The derivative of a function at a point gives the slope of the tangent line at that point.

Step 3 ：Since the given function is implicit, we will need to use implicit differentiation to find the derivative.

Step 4 ：Let's denote the function as \(f = -x e^{29 y} + y^{2} e^{\frac{58 x}{5}} + e^{2 x^{2} + 2 y^{2}}\).

Step 5 ：Using implicit differentiation, we find that \(\frac{df}{dx} = 4x e^{2 x^{2} + 2 y^{2}} + \frac{58 y^{2} e^{\frac{58 x}{5}}}{5} - e^{29 y}\) and \(\frac{df}{dy} = -29x e^{29 y} + 2y e^{\frac{58 x}{5}} + 4y e^{2 x^{2} + 2 y^{2}}\).

Step 6 ：The slope of the tangent line at the point (5,2) is given by \(-\frac{665}{327}\).

Step 7 ：Once we have the slope, we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form of a line is \(y - y_{1} = m(x - x_{1})\), where \(m\) is the slope and \((x_{1}, y_{1})\) is the given point.

Step 8 ：Substituting the slope and the given point into the point-slope form, we get the equation of the tangent line as \(y - 2 = -\frac{665}{327}(x - 5)\).

Step 9 ：Final Answer: The equation of the tangent line at the point (5,2) on the curve is \(\boxed{y - 2 = -\frac{665}{327}(x - 5)}\).