You can find the steps for balancing redox equations here.
Your unbalanced equation is
#"HNO"_3 + "C"_2"H"_6"O" + "K"_2"Cr"_2"O"_7 → "KNO"_3 + "C"_2"H"_4"O"+ "H"_2"O" + "Cr"("NO"_3)_3#
Step 1. Calculate the oxidation numbers of every atom.
Left hand side: #"H= +1; N= +5; O = -2; C = -2; K = +1; Cr = +6"#
Right hand side: #"K = +1; N = +5; O = -2; C = -1; H = +1; Cr = +3"#
Step 2. Calculate the changes in oxidation number:
#"C: -2 → -1; Change = +1"#
#"Cr: +6 → +3; Change = -3"#
Step 3. Balance changes in oxidation number.
You need 3 atoms of #"C"# for every 1 atom of #"Cr"#.
You have 2 atoms of #"Cr"#, so you need 6 atoms of #"C"#.
This gives you total changes of +6 and -6.
Step 4. Insert coefficients to get these numbers.
#"HNO"_3 + color(red)(3)"C"_2"H"_6"O" + color(blue)(1)"K"_2"Cr"_2"O"_7 → "KNO"_3 + color(red)(3)"C"_2"H"_4"O"+ "H"_2"O" + color(blue)(2)"Cr"("NO"_3)_3#
Step 5a. Balance #"K"#.
#"HNO"_3 + color(red)(3)"C"_2"H"_6"O" + color(blue)(1)"K"_2"Cr"_2"O"_7 → color(brown)(2)"KNO"_3 + color(red)(3)"C"_2"H"_4"O"+ "H"_2"O" + color(blue)(2)"Cr"("NO"_3)_3#
Step 5b. Balance #"N"#.
#color(green)(8)"HNO"_3 + color(red)(3)"C"_2"H"_6"O" + color(blue)(1)"K"_2"Cr"_2"O"_7 → color(brown)(2)"KNO"_3 + color(red)(3)"C"_2"H"_4"O"+ "H"_2"O" + color(blue)(2)"Cr"("NO"_3)_3#
Step 6. Balance #"O"#.
#color(green)(8)"HNO"_3 + color(red)(3)"C"_2"H"_6"O" + color(blue)(1)"K"_2"Cr"_2"O"_7 → color(brown)(2)"KNO"_3 + color(red)(3)"C"_2"H"_4"O"+ color(orange)(7)"H"_2"O" + color(blue)(2)"Cr"("NO"_3)_3#
Step 7. Balance #"H"#.
Done.
Step 8. Check that all atoms balance.
#"Atom"color(white)(m)"Left hand side"color(white)(m)"Right hand side"#
#color(white)(m)"H"color(white)(mmmmm)26color(white)(mmmmmmm)26#
#color(white)(m)"N"color(white)(mmmmmll)8color(white)(mmmmmmml)8#
#color(white)(m)"O"color(white)(mmmmm)34color(white)(mmmmmmm)34#
#color(white)(m)"C"color(white)(mmmmmll)6color(white)(mmmmmmml)6#
#color(white)(m)"Cr"color(white)(mmmmml)2color(white)(mmmmmmml)2#
The balanced equation is
#8"HNO"_3 + 3"C"_2"H"_6"O" + "K"_2"Cr"_2"O"_7 → 2"KNO"_3 + 3"C"_2"H"_4"O"+ 7"H"_2"O" + 2"Cr"("NO"_3)_3#