The problem is stated as follows:

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Show that the equation $\displaystyle zx^2y^3 + (x+y)e^{z-x}-x^2-x-4y=-3$ defines z as a function of x and y near (1,1,1). Also show that (1,1) is a stationary point of this function and determine its kind.

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The first part is easily shown with the implicit function theorem.

Differentiating the equation implicitly with respect to x:

$\displaystyle 2xy^3z + x^2y^3\frac{\partial z}{\partial x} + e^{z-x} + (\frac{\partial z}{\partial x} -1)(x+y)e^{z-x} -2x -1 = 0$, so $\displaystyle (x,y,z(x,y)) = (1,1,1)$ gives $\displaystyle 2 + \frac{\partial z}{\partial x} + 1 + 2(\frac{\partial z}{\partial x} - 1)-2-1=0$ near (x,y) = (1,1), and consequently, $\displaystyle \frac{\partial z}{\partial x} = \frac{2}{3}$ there.

Thus, (x,y) = (1,1) isn't a stationary point as the problem claims.

Feel free to point and laugh at me and tell me that (or preferably why) my arithmetic is wrong.