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(C) Given that, $u=e^{x^2-y^2}$.First, we find the partial derivative of $u$ with respect to $x$:$u_x = \frac{\partial}{\partial x}(e^{x^2-y^2}) = e^{

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$y u_x+x u_y=0$

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(C) Given that, $u=e^{x^2-y^2}$.First, we find the partial derivative of $u$ with respect to $x$:$u_x = \frac{\partial}{\partial x}(e^{x^2-y^2}) = e^{x^2-y^2}(2x)$.Multiplying by $y$, we get:$y u_x = 2xy e^{x^2-y^2} \quad (i)$.Next, we find the partial derivative of $u$ with respect to $y$:$u_y = \frac{\partial}{\partial y}(e^{x^2-y^2}) = e^{x^2-y^2}(-2y)$.Multiplying by $x$, we get:$x u_y = -2xy e^{x^2-y^2} \quad (ii)$.Adding equations $(i)$ and $(ii)$:$y u_x + x u_y = 2xy e^{x^2-y^2} - 2xy e^{x^2-y^2} = 0$.

If $u=e^{x^2-y^2}$, then

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