$\underline{\underline{x=1,y=2}}$  

$\to \frac{60-xy}{x^2+y^2}=\frac{60-1\cdot 2}{1^2+2^2}=\frac{60-2}{1+4}=\frac{58}{5}=11\frac{3}{5}$   

$\to \frac{y^2-17}{2x^2}+3x^{2}y=\frac{2^2-17}{2\cdot 1^2}+3\cdot 1^2\cdot 2=$  

$=\frac{4-17}{2\cdot 1}+3\cdot 1\cdot 2=-\frac{13}{2}+6=-6,5+6=-0,5$  

$\underline{\underline{x=2,y=\frac{1}{2}}}$  

$\to \frac{60-xy}{x^2+y^2}=\frac{60-2\cdot \frac{1}{2}}{2^2+{\left(\frac{1}{2}\right)}^2}=\frac{60-1}{4+\frac{1}{4}}=\frac{59}{4\frac{1}{4}}=\frac{59}{\frac{17}{4}}=59\cdot \frac{4}{17}=\frac{236}{17}=13\frac{15}{17}$  

$\to \frac{y^2-17}{2x^2}+3x^{2}y=\frac{{\left(\frac{1}{2}\right)}^2-17}{2\cdot 2^2}+3\cdot 2^2\cdot \frac{1}{2}=\frac{\frac{1}{4}-17}{2\cdot 4}+3\cdot 4\cdot \frac{1}{2}=$    

$=\frac{-16\frac{3}{4}}{8}+6=-\frac{67}{4}\cdot \frac{1}{8}+6=-\frac{67}{32}+6=-2\frac{3}{32}+6=3\frac{29}{32}$  

$\underline{\underline{x=-3,y=6}}$  

$\to \frac{60-xy}{x^2+y^2}=\frac{60-\left(-3\right)\cdot 6}{{\left(-3\right)}^2+6^2}=\frac{60-\left(-18\right)}{9+36}=\frac{60+18}{45}=\frac{78}{45}=1\frac{33}{45}$  

$\to \frac{y^2-17}{2x^2}+3x^{2}y=\frac{6^2-17}{2\cdot {\left(-3\right)}^2}+3\cdot {\left(-3\right)}^2\cdot 6=\frac{36-17}{2\cdot 9}+3\cdot 9\cdot 6=$  

$=\frac{19}{18}+162=1\frac{1}{18}+162=163\frac{1}{18}$  

$\underline{\underline{x=-0,5,y=1}}$  

$\to \frac{60-xy}{x^2+y^2}=\frac{60-\left(-0,5\right)\cdot 1}{{\left(-0,5\right)}^2+1^2}=\frac{60-\left(-0,5\right)}{0,25+1}=\frac{60,5}{1,25}=\frac{6050}{125}=48\frac{50}{125}=48\frac{2}{5}$      

$\to \frac{y^2-17}{2x^2}+3x^{2}y=\frac{1^2-17}{2\cdot {\left(-0,5\right)}^2}+3\cdot {\left(-0,5\right)}^2\cdot 1=\frac{1-17}{2\cdot 0,25}+3\cdot 0,5\cdot 1=$  

$=\frac{-16}{0,5}+1,5=-16\cdot 2+1,5=-32+1,5=-30,5$  

$x$	$1$	$2$	$-3$	$-0,5$
$y$	$2$	$\frac{1}{2}$	$6$	$1$
$\frac{60-xy}{x^2+y^2}$	$11\frac{3}{5}$	$13\frac{15}{17}$	$1\frac{33}{45}$	$48\frac{2}{5}$
$\frac{y^2-17}{2x^2}+3x^{2}y$	$-0,5$	$3\frac{29}{32}$	$163\frac{1}{18}$	$-30,5$