[tex]Well\;\mathrm{treat\:}y\mathrm{\:as\:}y\left(x\right) [/tex]
[tex]\mathrm{Differentiate\:both\:sides\:of\:the\:equation\:with\:respect\:to\:}x[/tex]
[tex]\frac{d}{dx}\left(2^xy+x^2-3xy\right)=\frac{d}{dx}\left(5\right)[/tex]
[tex]\frac{d}{dx}\left(2^xy+x^2-3xy\right) \ \textgreater \ \mathrm{Apply\:the\:Sum/Difference\:Rule}:\ \left(f\pm g\right)^'=f^'\pm g^'[/tex]
[tex]\frac{d}{dx}\left(2^xy\right) \ \textgreater \ \mathrm{Apply\:the\:Product\:Rule}:\ \left(f\cdot g\right)'=f^'\cdot g+f\cdot g^'\[/tex]
[tex]f=2^x,\:g=y \ \textgreater \ \frac{d}{dx}\left(2^x\right)y+\frac{d}{dx}\left(y\right)2^x[/tex]
[tex]\frac{d}{dx}\left(2^x\right) \ \textgreater \ \mathrm{Apply\:exponent\:rule}:\ \:a^b=e^{b\ln \left(a\right)} \ \textgreater \ 2^x=e^{x\ln \left(2\right)} [/tex]
[tex]\frac{d}{dx}\left(e^{x\ln \left(2\right)}\right) \ \textgreater \ \mathrm{Apply\:the\:chain\:rule}:\ \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}[/tex]
[tex]f=e^u,\:\:u=x\ln \left(2\right) \ \textgreater \ \frac{d}{du}\left(e^u\right)\frac{d}{dx}\left(x\ln \left(2\right)\right)[/tex]
[tex]\frac{d}{du}\left(e^u\right) \ \textgreater \ \mathrm{Apply\:the\:common\:derivative}:\ \frac{d}{du}\left(e^u\right)=e^u \ \textgreater \ e^u[/tex]
[tex]\frac{d}{dx}\left(x\ln \left(2\right)\right) \ \textgreater \ \mathrm{Take\:the\:constant\:out}:\ \left(a\cdot f\right)^'=a\cdot f^'[/tex]
[tex]\ln \left(2\right)\frac{d}{dx}\left(x\right) \ \textgreater \ \mathrm{Apply\:the\:common\:derivative}:\ \frac{d}{dx}\left(x\right)=1 \ \textgreater \ \ln \left(2\right)\cdot \:1[/tex]
[tex]Simplify \ \textgreater \ \ln \left(2\right)[/tex]
[tex]e^u\ln \left(2\right) \ \textgreater \ \mathrm{Substitute\:back}\:u=x\ln \left(2\right) \ \textgreater \ e^{x\ln \left(2\right)}\ln \left(2\right) [/tex]
[tex]e^{x\ln \left(2\right)} \ \textgreater \ \mathrm{Apply\:exponent\:rule}:\ \:a^{bc}=\left(a^b\right)^c \ \textgreater \ \left(e^{\ln \left(2\right)}\right)^x[/tex]
[tex]\mathrm{Apply\:log\:rule}:\ \:a^{\log _a\left(b\right)}=b \ \textgreater \ e^{\ln \left(2\right)}=2 \ \textgreater \ \ln \left(2\right)\cdot \:2^x[/tex]
[tex]\ln \left(2\right)\cdot \:2^xy+\frac{d}{dx}\left(y\right)2^x[/tex]
[tex]\frac{d}{dx}\left(x^2\right) \ \textgreater \ \mathrm{Apply\:the\:Power\:Rule}:\ \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1} \ \textgreater \ 2x^{2-1} \ \textgreater \ 2x[/tex]
[tex]\frac{d}{dx}\left(3xy\right) \ \textgreater \ \mathrm{Take\:the\:constant\:out}:\ \left(a\cdot f\right)^'=a\cdot f^'[/tex]
[tex]3\frac{d}{dx}\left(xy\right) \ \textgreater \ \mathrm{Apply\:the\:Product\:Rule}:\ \left(f\cdot g\right)'=f^'\cdot g+f\cdot g^'[/tex]
[tex]f=x,\:g=y \ \textgreater \ 3\left(\frac{d}{dx}\left(x\right)y+\frac{d}{dx}\left(y\right)x\right)[/tex]
[tex]\frac{d}{dx}\left(x\right) \ \textgreater \ \mathrm{Apply\:the\:common\:derivative}:\ \frac{d}{dx}\left(x\right)=1 [/tex]
[tex]3\left(1\cdot \:y+\frac{d}{dx}\left(y\right)x\right) \ \textgreater \ Simplify \ \textgreater \ 3\left(y+x\frac{d}{dx}\left(y\right)\right)[/tex]
[tex]\frac{d}{dx}\left(5\right) \ \textgreater \ \mathrm{Derivative\:of\:a\:constant}:\ \frac{d}{dx}\left(a\right)=0 [/tex]
[tex]Now\;we\;have \ \textgreater \ \ln \left(2\right)\cdot \:2^xy+\frac{d}{dx}\left(y\right)2^x+2x-3\left(y+x\frac{d}{dx}\left(y\right)\right)[/tex]
[tex]\mathrm{For\:convenience,\:write\:}\frac{d}{dx}\left(y\right)\mathrm{\:as\:}y^{'\:} [/tex]
[tex]\ln \left(2\right)\cdot \:2^xy+y^{'\:}\cdot \:2^x+2x-3\left(y+xy^{'\:}\right)=0[/tex]
Now we have to isolate y^'.
[tex]\ln \left(2\right)\cdot \:2^xy+y^{'\:}\cdot \:2^x+2x-3\left(y+xy^{'\:}\right)=0 [/tex]
[tex]\mathrm{Subtract\:}\ln \left(2\right)\cdot \:2^xy+2x\mathrm{\:from\:both\:sides}[/tex]
Too big for the formula so i assume you know how to subtract them.
Simplify.
[tex]y^{'\:}\cdot \:2^x-3\left(y+xy^{'\:}\right)=-\left(\ln \left(2\right)\cdot \:2^xy+2x\right) \ \textgreater \ expand -3\left(y+xy^{'\:}\right)[/tex]
[tex]\mathrm{Distribute\:parentheses\:using}:\ \:a\left(b+c\right)=ab+ac [/tex]
[tex]where\; a=-3,\:b=y,\:c=xy^{'\:}[/tex]
[tex]-3\cdot \:y-3\cdot \:xy^{'\:} \ \textgreater \ y^{'\:}\cdot \:2^x-3y-3xy^{'\:}=-\left(\ln \left(2\right)\cdot \:2^xy+2x\right)[/tex]
[tex]Expand\; -\left(\ln \left(2\right)\cdot \:2^xy+2x\right) \ \textgreater \ \mathrm{Distribute\:parentheses}[/tex]
[tex]-\ln \left(2\right)\cdot \:2^xy-2x \ \textgreater \ \mathrm{Apply\:minus-plus\:rules} \ \textgreater \ +\left(-a\right)=-a[/tex]
[tex]-\ln \left(2\right)\cdot \:2^xy-2x \ \textgreater \ 2^xy^{'\:}-3y-3xy^{'\:}=-\ln \left(2\right)\cdot \:2^xy-2x[/tex]
[tex]\mathrm{Add\:}3y\mathrm{\:to\:both\:sides} [/tex]
[tex]2^xy^{'\:}-3y-3xy^{'\:}+3y=-\ln \left(2\right)\cdot \:2^xy-2x+3y[/tex]
[tex]\mathrm{Simplify} \ \textgreater \ 2^xy^{'\:}-3xy^{'\:}=-\ln \left(2\right)\cdot \:2^xy-2x+3y[/tex]
[tex]2^xy^{'\:}-3xy^{'\:} \ \textgreater \ \mathrm{Factor\:out\:common\:term\:}y^{'\:} \ \textgreater \ y^{'\:}\left(-3x+2^x\right)[/tex]
[tex]y^{'\:}\left(-3x+2^x\right)=-\ln \left(2\right)\cdot \:2^xy-2x+3y [/tex]
