Use an orthogonal transformation and a translation to express in standard form the curve given by the equation
This probably isn't the way you're meant to do the question, but I can't help noticing that this equation factorises as $\displaystyle (x-3y)(x-3y+2\sqrt{10})=30$, and that it then factorises further as $\displaystyle (x-3y-\sqrt{10})(x-3y+3\sqrt{10}) = 0.$ This is the equation of two parallel straight lines with slope 1/3. If you rotate the axes through an angle of -arctan(1/3) they become the lines x=1 and x=–3, and if you then translate -1 unit in the direction of the x-axis, they become x=±2. That looks like the standard form of this "curve".