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- Thread starter jaguar7
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[tex]\frac{x^2}{4}+ y^2+ z^2= 3[/tex]

is not a function, it is an equation. If you mean that equation defines z implicitely as a function of x and y, then, differentiating with respect to x, [itex]x/2+ 2z z_x= 0[/itex] so that [itex]z_x= -x/4z[/itex] and [itex]2y+ 2z z_y= 0[/itex] so that [itex]z_y= -y/z[/itex]. Notice that, in the first case, the derivative of y with respect to x is 0 and, in the second, the derivative of x with resepect to y is 0. That is because they are independent variables.

Of course, we could just as easily think of that equation as defining y in terms of the variables x and z and have [itex]x/2+ 2yy_x= 0[/itex] and [itex]2y y_z+ 2z= 0[/itex] or we could think of it as defining x in terms of y and z so that [itex](x/2)x_y+ 2y= 0[/itex] and [itex](x/2)x_z+ 2z= 0[/itex].

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