Look at it this way: You have three variables $u$, $x$, $y$ (modeling, e.g., physical quantities) that are related by the "constituent equation"
$$\bigl(F(x,y,u):=\bigr)\qquad x^u+u^y-u=0\ .\tag{1}$$
An admissible state is, e.g., the point ${\bf p}_0:=(1,0,2)$. The equation $(1)$ reduces the number of degrees of freedom from $3$ to $2$; therefore we may nourish the idea that given $x$ and $y$ the value of $u$ is determined. In other words: There is an underlying function
$$\phi:\quad(x,y)\mapsto u=\phi(x,y)\ ,$$
which produces for all suitable $(x,y)$ an $u$ such that the triple ${\bf p}:=\bigl(x,y,u\bigr)$ is admissible. This means that
$$F\bigl(x,y,\phi(x,y)\bigr)=0\quad\forall x,\ \forall y\ .$$
Taking partial derivatives with respect to $x$ and $y$, and using the chain rule we can deduce that
$$F_x\bigl(x,y,\phi(x,y)\bigr)+F_u\bigl(x,y,\phi(x,y)\bigr)\phi_x(x,y)\equiv0,\quad
F_y\bigl(x,y,\phi(x,y)\bigr)+F_u\bigl(x,y,\phi(x,y)\bigr)\phi_y(x,y)\equiv0\ .$$
Solving for the partial derivatives of $\phi$ we get
$$\phi_x(x,y)=-{F_x\bigl(x,y,\phi(x,y)\bigr)\over F_u\bigl(x,y,\phi(x,y)\bigr)}, \quad
\phi_y(x,y)=-{F_y\bigl(x,y,\phi(x,y)\bigr)\over F_u\bigl(x,y,\phi(x,y)\bigr)}\ .$$
This can be interpreted as follows: Near any admissible point ${\bf p}_0=(x_0,y_0,u_0)$ the "local" function $(x,y)\mapsto u:=\phi(x,y)$ has partial derivatives
$${\partial u\over\partial x}=-{F_x({\bf p})\over F_u({\bf p})}, \quad
{\partial u\over\partial y}=-{F_y({\bf p})\over F_u({\bf p})}\ ,\tag{2}$$
under the essential condition that $F_u({\bf p}_0)\ne0$.
In our example $F_u(x,y,u)=\log x\cdot x^u- y u^{y-1}-1$; so at the point ${\bf p}_0:=(1,0,2)$ we have $F_u({\bf p}_0)=-1$. Therefore the formula $(2)$ is applicable at ${\bf p}_0$. I leave it to the OP to compute the values of the partial derivatives appearing therein.
