Given equation of curve is "y=3x^2-1"
"\\therefore \\frac{dy}{dx}=6x"
So, slope of tangent at any point "(x_1,y_1)" is: "6x_1"
Now, according to question, this tangent is parallel to the given line "2x-y+3=0"
So, "6x_1=2\\\\"
"\\Rightarrow x_1=\\frac{1}{3}"
"\\therefore y_1=3x_1^2-1=3(\\frac{1}{3})^2-1=-\\frac{2}{3}"
So, equation of tangent is: "y+\\frac{2}{3}=2(x-\\frac{1}{3})\\\\"
"\\Rightarrow 3y+2=6x-2\\\\\n\\Rightarrow 6x-3y-4=0"
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