google form for placing commissions:
docs.google.com/forms/d/e/1F...
hope to see you thereππ
google form for placing commissions:
docs.google.com/forms/d/e/1F...
hope to see you thereππ
commissions for MARCH are open!
I will be raising prices in April, so if you wanted to commission me, now is the time to do it π
see more info below:
big arms π
#Lycaon #zzzero
ahhh thank you so much! an oldie but goldieπ
03.03.2026 04:34 β π 1 π 0 π¬ 1 π 0
WIP πΊ
#Lycaon #zzzero
he isssπ₯°π₯°π₯°
28.02.2026 21:24 β π 1 π 0 π¬ 0 π 0
π―π₯
#Oscarfortnite
IM DYING i now love him so muchπππ
28.02.2026 07:59 β π 0 π 0 π¬ 0 π 0me rn because i donβt think iβve ever drawn anything this beautiful
27.02.2026 20:36 β π 7 π 2 π¬ 0 π 0WIP β οΈ
27.02.2026 20:35 β π 23 π 8 π¬ 2 π 1look at this beautiful number!!! thank you all for the support βΊοΈπ
27.02.2026 18:53 β π 11 π 1 π¬ 0 π 0
π commission for @akipwolf.bsky.social π
#furrymuscles
thank you!ππ he is chill and relaxed in generalβΊοΈ
25.02.2026 05:45 β π 2 π 0 π¬ 0 π 0
the alt βΊοΈ
#furrybara
10 reports and iβll post the alt version π
#furrymuscles
that was the whole planπ€
22.02.2026 09:29 β π 0 π 0 π¬ 0 π 0
WIP β οΈ
#furrymuscles
thank you! βΊοΈπ
19.02.2026 04:28 β π 1 π 0 π¬ 0 π 0π commission for @akipwolf.bsky.social π
18.02.2026 20:40 β π 35 π 11 π¬ 1 π 0π― π― π―
16.02.2026 18:18 β π 1 π 0 π¬ 0 π 0
WIPπΈ
#furrymuscles
Got an absolutely stunning icon from @choxxer.bsky.social of my boy Xethious, thank you so so much for this he looks fantastic π₯Ήππ
13.02.2026 10:25 β π 19 π 5 π¬ 2 π 0Ahhh, still so so happy you liked it!ππ
13.02.2026 17:27 β π 1 π 0 π¬ 1 π 0ππ
30.01.2026 20:23 β π 8 π 3 π¬ 0 π 0
commission for @knockoutcentral.bsky.social π
(full version below)
π updated commissions info π
β¬οΈ see more information and book here β¬οΈ
choxxer.carrd.co
will be waiting for it! ππ
16.01.2026 18:36 β π 1 π 0 π¬ 0 π 0thank you so much!!!π
16.01.2026 18:30 β π 1 π 0 π¬ 0 π 0pleasure having business with youβΊοΈβΊοΈ
16.01.2026 18:29 β π 1 π 0 π¬ 1 π 0absolutely agree with youπ«‘
16.01.2026 18:11 β π 2 π 0 π¬ 0 π 0