So I have this system
$$\frac{dx}{dt} = x y$$
$$\frac{dy}{dt} = 2 y$$
$$(x(0),y(0)) = (1,1)$$
Although I'm not too sure where to start. I know one method you have to take the derivative with respect to t for one of them and eliminate the other through substituting the $x'(t)$ or $y'(t)$ into it and using regular ODE solving techniques.
Just like Winther mentioned, start by solving $y'=2y$, for example use the method of separation of variables:
$$\frac{dy}{dt}=2y \Leftrightarrow \int_1^y \frac{1}{y} dy = \int_0^t 2 dt \Leftrightarrow y=e^{2t}$$
And now the first ODE $x'=xy=xe^{2t}$ and also by the same method:
$$ \frac{dx}{dt} = x e^{2t} \Leftrightarrow \int_1^x \frac{1}{x} dx = \int_0^t e^{2t} dt \Leftrightarrow x=e^{\frac{1}{2} (e^{2t}-1) } $$
Notice, solving second differential equation $$\frac{dy}{dt}=2y$$ $$\int \frac{dy}{y}=2\int \ dt$$ $$\ln y=2t+c$$ setting $y=1$ at $t=0\implies c=0$, hence, $$\ln y=2t$$ $$\color{red}{y=e^{2t}}$$ setting value of $y$ in first differential equation, $$\frac{dx}{dt}=xy=xe^{2t}$$ $$\frac{dx}{x}=e^{2t}\ dt$$ $$\int \frac{dx}{x}=\int e^{2t}\ dt$$ $$\ln x=\frac 12e^{2t}+C$$ setting $x=1$ at $t=0\implies C=-\frac 12$, hence $$\ln x=\frac{e^{2t}}{2}-\frac 12=\color{}{\frac{1}{2}(e^{2t}-1)}$$ $$\color{red}{x=e^{\Large \frac{1}{2}(e^{2t}-1)}}$$
$$y'(t)=2y(t)\Longleftrightarrow$$ $$\frac{y'(t)}{y(t)}=2\Longleftrightarrow$$ $$\int\frac{y'(t)}{y(t)}\space\text{d}t=\int2\space\text{d}t\Longleftrightarrow$$ $$\ln|y(t)|=2t+\text{C}\Longleftrightarrow$$ $$y(t)=\text{C}e^{2t}$$
$$x'(t)=x(t)y(t)\Longleftrightarrow$$ $$x'(t)=x(t)\text{C}e^{2t}\Longleftrightarrow$$ $$\frac{x'(t)}{x(t)}=\text{C}e^{2t}\Longleftrightarrow$$ $$\int\frac{x'(t)}{x(t)}\space\text{d}t=\int\text{C}e^{2t}\space\text{d}t\Longleftrightarrow$$ $$\ln|x(t)|=\frac{\text{C}e^{2t}}{2}+\text{K}\Longleftrightarrow$$ $$x(t)=e^{\frac{\text{C}e^{2t}}{2}+\text{K}}$$
