Answer
$$\frac{\partial f(1,1)}{\partial x}=\frac{10}{27}$$
$$\frac{\partial f(1,1)}{\partial y}=\frac{8}{27}$$
Work Step by Step
We will use quotient rule to find both partial derivatives.
The partial derivative with the respect to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}\left(\frac{2xy}{\sqrt{4x^2+5y^2}}\right)=\frac{\sqrt{4x^2+5y^2}\frac{\partial }{\partial x}(2xy)-2xy\frac{\partial }{\partial x}(\sqrt{4x^2+5y^2})}{(\sqrt{4x^2+5y^2})^2}=\frac{\sqrt{4x^2+5y^2}\cdot2y-2xy\cdot\frac{1}{2\sqrt{4x^2+5y^2}}\frac{\partial }{\partial x}(4x^2+5y^2)}{4x^2+5y^2}=\frac{(4x^2+5y^2)\cdot2y-xy\cdot8x}{(\sqrt{4x^2+5y^2})^3}=\frac{8x^2y+10y^3-8x^2y}{(\sqrt{4x^2+5y^2})^3}=\frac{10y^3}{(\sqrt{4x^2+5y^2})^3}$$
because $y$ is treated as a constant. We alse used chain rule to find $\frac{\partial }{\partial x}(\sqrt{4x^2+5y^2})$.
The partial derivative with the respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}\left(\frac{2xy}{\sqrt{4x^2+5y^2}}\right)=\frac{\sqrt{4x^2+5y^2}\frac{\partial }{\partial y}(2xy)-2xy\frac{\partial }{\partial y}(\sqrt{4x^2+5y^2})}{(\sqrt{4x^2+5y^2})^2}=
\frac{\sqrt{4x^2+5y^2}\cdot2x-2xy\cdot\frac{1}{2\sqrt{4x^2+5y^2}}\frac{\partial }{\partial y}(4x^2+5y^2)}{4x^2+5y^2}=\frac{(4x^2+5y^2)\cdot2x-xy\cdot10y}{(\sqrt{4x^+5y^2})^3}=\frac{8x^3+10y^2x-10y^2x}{(\sqrt{4x^2+5y^2})^3}=\frac{8x^3}{(\sqrt{4x^2+5y^2})^3}$$
because now $x$ is treated as a constant. Again, we used the chain rule to find $\frac{\partial}{\partial y}(\sqrt{4x^2+5y^2}).$
Now we will evaluate these partial derivatives in the given point $(x,y)=(1,1)$:
$$\frac{\partial f(1,1)}{\partial x}=\left.\frac{10y^3}{(\sqrt{4x^2+5y^2})^3}\right|_{(x,y)-(1,1)}=\frac{10\cdot1^3}{(\sqrt{4\cdot1^2+5\cdot1^2})^3}=\frac{10}{(\sqrt9)^3}=\frac{10}{27}$$
$$\frac{\partial f(1,1)}{\partial y}=\left.\frac{8x^3}{(\sqrt{4x^2+5y^2})^3}\right|_{(x,y)=(1,1)}=\frac{8\cdot1^3}{(\sqrt{4\cdot1^2+5\cdot1^2})^3}=\frac{8}{(\sqrt9)^3}=\frac{8}{27}$$