Answer
The partial derivatives
$$\frac{\partial f(x,y)}{\partial x}$$ and $$\frac{\partial f(x,y)}{\partial y}$$
are not defined in the given point $(x,y)=(1,1).$
Work Step by Step
To solve both partial derivative we will use chain rule.
The partial derivative with respcet to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(\arccos xy)=-\frac{1}{\sqrt{1-(xy)^2}}\cdot\frac{\partial }{\partial x}(xy)=-\frac{y}{\sqrt{1-(xy)^2}}$$
because $y$ is treated as a constant.
The partial derivative with the respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(\arccos xy)=-\frac{1}{\sqrt{1-(xy)^2}}\frac{\partial }{\partial y}(xy)=-\frac{x}{\sqrt{1-(xy)^2}}$$
because here $x$ is treated as a constant.
Now we will evaluate these partial derivatives in the given point $(x,y)=(1,1)$:
$$\frac{\partial f(1,1)}{\partial x}=\left.-\frac{y}{\sqrt{1-(xy)^2}}\right|_{(x,y)=(1,1)}=-\frac{1}{\sqrt{1-1\cdot1}}=-\frac{1}{0}$$
We see that $0$ is in the denominator, so because we cannot divide with $0$ the partial derivative with respect to $x$ is not defined at this point.
$$\frac{\partial f(1,1)}{\partial y}=\left.-\frac{x}{\sqrt{1-(xy)^2}}\right|_{(x,y)=(1,1)}=-\frac{1}{\sqrt{1-1\cdot1}}=-\frac{1}{0}$$
We see that $0$ is in the denominator, so because we cannot divide with $0$ the partial derivative with respect to $y$ is not defined at this point.