Interchanging partial differentiation with evaluation along a curve

Question

Suppose I have a continuously differentiable function of three variables, $\gamma(x, y, z)$. Is it always valid to interchange the order of the operations of evaluation at a point, and differentiation? That is, is it necessarily true that $$ \left[\frac{\partial \gamma(x, y, z)}{\partial x}\right]_{z=x} = \frac{\partial}{\partial x}[\gamma(x, y, x)]\tag{1} $$ and $$ \left[\frac{\partial \gamma(x, y, z)}{\partial y}\right]_{z=x} = \frac{\partial}{\partial y}[\gamma(x, y, x)]?\tag{2} $$ In particular, if there is an additional constraint on the function so that $$ \gamma(x, y, x) = y, $$ is it true that $$ \left[\frac{\partial \gamma(x, y, z)}{\partial y}\right]_{z=x} = \frac{\partial}{\partial y}[\gamma(x, y, x)] = \frac{\partial}{\partial y}[y] = 1,\tag{3} $$ and that $$ \left[\frac{\partial \gamma(x, y, z)}{\partial x}\right]_{z=x} = \frac{\partial}{\partial x}[\gamma(x, y, x)] = \left[\frac{\partial \gamma(x, y, z)}{\partial z}\right]_{z=x}? \tag{4} $$ For context, see my question over on Physics SE. Thanks!

Answer 1

No, plenty of counterexamples: Let $\gamma (x,y,z) = x+y-z.$ Then

$$\frac{\partial \gamma}{\partial x}(x,y,z) = 1,$$

whether $x=z$ or not. But

$$\frac{\partial \gamma}{\partial x}(x,y,x) = \frac{\partial }{\partial x}\,y =0.$$